3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 7 InterviewSolution

301.	Cell reactiomn is spontaneous whenA. `E_"red"^@` is negativeB. `DeltaG^@` is negativeC. `E_"oxide"^@` is positiveD. `DeltaG^@` is positive
Answer» Correct Answer - B

Discussion

302.	Mark the correct Nernst equation for the given cell. `F_((s))\|Fe^(2+)(0.001 M) \|\|H^(+)(1M) \| H_(2(g)) ( 1 bar ) \| Pt_((s))` isA. `E_"cell"=E_"cell"^@-"0.591"/2 "log" ([Fe^(2+)][H^+]^2)/([Fe][H_2])`B. `E_"cell"=E_"cell"^@-"0.591"/2 "log" ([Fe][H^+]^2)/([Fe^(2+)][H_2])`C. `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([Fe^(2+)][H_2])/([Fe][H^+]^2)`D. `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([Fe][H_2])/([Fe^(2+)][H^+]^2)`
Answer» Correct Answer - C `{:("At anode:",Fe to Fe^(2+) (0.001 M)+2e^-),("At cathode:",2H^(+)(1M)+2e^(-) to H_2"(1 atm)"),("Net reaction:",Fe+2H^(+) to Fe^(2+)+ H_2):}` Nernst equation for the given cell , `E_"cell"=E_"cell"^@-0.0591/2 "log"([Fe^(2+)][H_2])/([Fe][H^+]^2)`

Discussion

303.	The correct Nernst equation for the given cell `Pt_((s)) \| Br_(2(l)) \| Br^(-) (M) \|\| H^(+) (M) \|H_(2(g)) (1bar) \| Pt_((s))` isA. `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([Br_(2(l))][H_2])/([H^+]^2[Br^-]^2)`B. `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([H^+]^2[Br^-]^2)/([Br_(2(l))][H_2])`C. `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([H^+]^2[H_2])/([Br_(2(l))][Br^(-)]^2)`D. `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([Br_(2(l))][Br^-]^2)/([H^+]^2[H_2])`
Answer» Correct Answer - A

Discussion

304.	At `25^(@)C` , the free energy of formation of `H_(2)O(l)` is `-56,700 Cal mol^(-1)`. The free energy of ionization of water to `H^(o+)` and `overset(c-)(O)H` is `19050 cal mol^(-1)`. What is the reversible `EMF` of the following cell at `25^(@)C:` `H_(2)(g)(1 atm)\|H^(o+)\|\|overset(c-)(O)H,O_(2)(g),1atm`
Answer» Correct Answer - `0.405 V ;`

Discussion

305.	Find the reduction potential of `AsO_(4)^(3-)\|AsO_(2)^(c-)` in a solution when `18mL` of `0.1 N` solution of `NaI` is added to `20mL` of `0.1N Na_(3)AsO_(4)` solution at `pH=5.` The standard reduction potential of `AsO_(4)^(3-)\|AsO_(2)^(c-)=-0.70V`.
Answer» Half cell reaction `:` `4H^(o+)+AsO_(4)^(3-)+2e^(-) rarr AsO_(2)^(c-)+2H_(2)O` `E_((AsO_(4)^(3-)\|AsO_(4)^(3-)))=E^(c-)._((AsO_(4)^(3-)\|AsO_(2)^(c-)))` `-(0.059)/(2) log .([AsO_(2)^(c-)])/([AsO_(4)^(3-)][H^(o+)]^(4))` `mEq ` of `AsO_(4)^(3-)=20xx0.1=2` `mEq` of `I^(c-)=18xx0.1=1.8` Therefore, `1.8 m Eq` of `I^(c-)` reacts with `1.8 mEq` of `AsO_(4)^(3-)` to give `1.8 mEq` of `AsO_(2)^(c-)` So `mEq` of `AsO_(4)^(3-)` left `2-1.8=0.2 meQ`. Total volume of solution `=(18+20)mL=38mL` Hence, `[AsO_(4)^(3-)]=(0.2mEq)/(38mL),[AsO_(2)^(c-)]=(1.8mEq)/(38mL)` `pH=5implies[H^(o+]=10^(-5)M` Substituting the values in Eq. `(i)` `E_((AsO_(4)^(3-)\|AsO_(2)^(c-)))=E^(c-)._((AsO_(4)^(3-)\|AsO_(2)^(c-)))` `-(0.059)/(2) log . ([AsO_(2)^(c-)])/([AsO_(4)^(3-)][H^(o+)]^(4))` `=-0.70V-(0.059)/(2)log .((1.8//38))/(((0.2)/(38))(10^(-5))^(4))` `=-0.70V-(0.059)/(2)log .((1.8)/(0.2xx10^(-20)))` `=-0.70V-(0.059)/(2)[log 9 + log 10^(20)]` `=-0.70V-(0.059)/(2)[2log3+ 20] (Take 0.059=0.06log3~~0.48)` `=-0.70V-0.30[2xx0.48+20]` `=-0.70V-0.03xx20.96` `=(-0.7-0.6288)V=-1.328V` `:. `Hence,`E_((AsO_(4)^(3-)\|AsO_(2)^(c-)))=-1.328V`

Discussion

306.	The following galvanic cell was `Zn\|{:(Zn(NO_(3))_(2(aq.))),(100 mL"," 1M):}\|\|{:(Cu(NO_(3))_(2(aq.))),(100 mL"," 1M):}\|Cu` operated as ann electrolytic cell using `Cu` as anode `Zn` as cathode. A current of `0.48` ampere was passed for `10` hour and then the cell was allowed to funcation as galvanic cell. What would be the `e.m.f.` of the cell at `25^(@)C`? Assume that the only electrode reactions occuring were those involving `Cu//Cu^(2+)` and `Zn//Zn^(2+)`. (Given `E_(Cu^(3+)//Cu)^(@) = + 0.34 V` and `E_(Zn^(2+)//Zn)^(@) = -0.76 V`)
Answer» Correct Answer - `1.137 V ;`

Discussion

307.	A constant current was passed through a solution of `AuCl_(4)^(c-)` ion between gold electrodes. After a period of `10.0 mi n`, the increase in the weight of cathode was `1.314g`. The total charge passed through solution is `(` atomic weight of `AuCl_(4)^(c-)=339)`A. `1.16 xx 10^(-2)F`B. `3.5xx10^(-2)F`C. `2xx10^(-2)F`D. `4xx10^(-3)F`
Answer» Correct Answer - c `Au^(3+)+3e^(-) rarr Au` `(W)/(197//3)=(Charg e)/(96500)` `:. Charg e=(1.314xx3xxF)/(197)=0.02F`

Discussion

308.	If in a galvanic cell, the cell reaction is reversed as `Cu(s)\|Cu^(2_(aq)\|\|Zn^(2)aq\|Zn` the cell potential will beA. 11.1VB. `-11.1V`C. `-1.1V`D. `1.1V`
Answer» Correct Answer - c `Cu^(2+)+2e^(-)rarrCu,E^(@)=+0.337V` `Zn^(2)++2e^(-)rarrZn,E^(@)=-0.763V` In the cell `Cu\|Cu^(2+)\|\|Zn^(2+)\|\|Zn` `E_("cell")=E_("cathode")^(@)-E_("anode")^(@)` `=-0.763(+0.337)=-1.10`

Discussion

309.	Calculate the potential of the following cell `:` `Pt(s)\|{:(Ce^(3+),(2M)) ,(Ce^(4+),(1M)):}\|\|{:(Cr^(3+)(2M),) ,(Cr_(2)O_(7)^(2-)(1M),H^(o+)(1M)):}\|Pt(s)` Given `:E^(c-)._(Ce^(3+)\|Ce^(4+))=-1.7,E^(c-)._(Cr_(2)O_(7)^(2-)\|Cr^(3+))=1.3V` `(`Take`0.059~~0.06)`
Answer» `E^(c-)._(Ce^(3+)\|Ce^(4+))=-1.7V` `(` Standard oxidation potential value `)`. So, `E^(c-)._(Ce^(4+)\|Ce^(3+))=1.7V` `(` Standard reduction potential value `).` and `E^(c-)._((Cr_(2)O_(7)^(2-)\|Cr^(3+)))=1.3V` `(` Standard reduction potential value `)`. Anode reduction `:` `6Ce^(3+)(2M) rarr 6Ce^(4+)(1M)+cancel(6e^(-))` Cathode reaction `:` `14H^(o+)(1M)+Cr_(2)O_(7)^(2-)(2M)+cancel(6e^(-))rarr2Cr^(3+)(1M)+7H_(2)O` Cell reaction `:` `ulbar(6Ce^(3+)(2M)+14H^(o+)(1M)+Cr_(2)O_(7)^(2-)(2M)rarr 6Ce^(4+)(1M)+7H_(2)O)` `E^(c-)._(cell)=(E^(c-)._(reduction))_(c)-(E^(c-)._(reduction ))_(a)` `=1.3V-1.7V=-0.4V` `E_(cell)=E^(c-)._(cell)-(0.06)/(6)log .([Ce^(4+)]^(6)[Cr^(3+)]^(2))/([Ce^(3+)]^(6)[H^(o+)]^(14)[Cr_(2)O_(7)^(2-)])` `=-0.4V-0.01log.((1)^(6)(1)^(2))/((2)^(6)(1)^(14)(2))` `=-0.4V-0.01(log2^(-7))` `=-0.4V-0.01(-7 log 2)` `=(-0.4+0.01xx7xx0.3)V` `=(-0.4+0.021)V=-0.379~~-0.38V`

Discussion

310.	Calculate the potential of the following cell `:` `Pt\|underset((2.0M)(1.0M))(Co^(2+),Co^(3+))\|underset((1.0M)(4.0M)(1.0M))(Cr^(3+),Cr_(2)O_(7)^(2_-),H^(o+))\|Pt` `E^(c-)._(Co^(2+)\|Co^(3+))=-2V,E^(c-)._(Cr_(2)O_(7)^(2-)\|Cr^(3+))=+1.0V`A. `1.024V`B. `-1.024V`C. `0.976V`D. `-0.976V`
Answer» Correct Answer - d `Co^(2+) rarr Co^(3+)+e^(-)]xx6` `14H^(o+)+6e^(-)+Cr_(2)O_(7)^(2-)rarr2Cr^(3+)+7H_(2)O` `ulbar(6Co^(2+)+14H^(o+)+Cr_(2)O_(7)^(2-)rarr6Co^(3+)+2Cr^(3+)+7H_(2)O)` `E^(c-)._(cell)=E^(c-)._(c)-E^(c-)._(a)(E^(c-)._(red)Co^(3+)\|Co^(2+)=2V)` `=1-2=-1V` `E=E^(c-)-(0.059)/(6)log.([Co^(3+)]^(6)[Cr^(3+)]^(2))/([Co^(2+)]^(6)[H^(o+)]^(14)[Cr_(2)O_(7)^(2-)])` `=-1-0.01log.((1)^(6)xx(1)^(2))/((2)^(6)xx(1)^(14)xx(4))` `=-1-0.01log(2)^(-8)` `=-1+0.01xx8xx0.3` `=-0.976V`

Discussion

311.	Calculate the potential corresponding to the following cell. Given `Pt\|Co^(2+)(2.0M),Co^(3+)(0.01M)\|\|Cr^(3+)(0.5M),Cr_(2)O_(7)^(2-)(4.0M),H^(o+)(1.5M)\|Pt ` `E^(c-)._(Co^(2+)\|Co^(3+))=-1.82V` `E^(c-)._(Cr_(2)O_(7)^(2-)\|Cr^(3+))=+1.33V`
Answer» `Co^(2+)rarr Co^(3+)+e^(-)]xx6` `14H^(o+) +Cr_(2)O_(7)^(2-)+6e^(-)rarr2Cr^(3+)+7H^(2)O` `ulbar(6Co^(+2)+14H^(o+)+Cr_(2)O_(7)^(2-)rarr6Co^(+3)+2Cr^(3+)+7H_(2)O)` `E^(c-)._(cell)=-1.82+1.33V=-0.49` `E_(cell)=E^(c-)._(cell)-(0.0591)/(6)log.([Co^(3+)]^(6)[Cr^(3+)]^(2))/([Co^(2+)]^(6)[Cr_(2)O_(7)^(2-)][H^(o+)]^(14))` `=-0.49-(0.0591)/(6)log.([10^(-2)]^(6)[0.50]^(2))/([2.0]^(6)[4.0][1.5]^(14))=-0.32V`

Discussion

312.	Calculate the reduction potential of the following electrodes `:` `a.` `Pt,H_(2)(4 atm)\|H_(2)SO_(4)(0.01M)` `b.` `Pt,H_(2)(1 atm)\| HCl (0.2 M)` `c.` Calculate the potential of hydrogen electrode in contact with a solution whose `i pH=5" "ii. pOH=4`
Answer» `a.` Half cell reaction `:(n` factor for `H_(2)SO_(4)=2)` `H_(2)(4 atm)rarr 2H^(o+)(0.01M o r 0.02 N) +2e^(-)` `E_((H_(2)\|2H^(o+)))=E^(c-)._((H_(2)\|2H^(o+)))-(0.059)/(2)log.([H^(o+)])/(p_(H_(2)))` `0-(0.059)/(2)log.((0.02)^(2))/(4 atm)` `=-(0.059)/(2)log((0.02xx0.02)/(4))` `=-(0.059)/(2)log(10^(-4)) ` `=-(0.59)/(2)xx-4log 10` `=0.059xx2=0.118V` `E_((2H^(o+)\|H_(2)))=-0.118V` `b.` Half cell reaction `:(n` factor for `HCl=1)` `H_(2)(1 atm) rarr2H^(o+)(0.2M o r 0.2 N)+2e` `E_((H_(2)\|2H^(o+)))=E^(c-)._((H_(2)//2H^(o+)))-(0.059)/(2)log.([H^(o+)]^(2))/([p_(H_(2))])` `=0-(0.059)/(2)log.((0.2xx0.2))/(1 atm)` `=-(0.059)/(2)[ log 4xx10^(-2))]` `=-(0.059)/(2)[2 log 2-2]` `=(0.059)/(2)xx-1.4=0.059xx0.7=0.0413V` `E_((2H^(o+)\|H_(2)))=-0.0413V` `c.` `i.` Use direct formula for the potential `(EMF)` of hydrogen electrode. `E_((EH^(o+)\|H_(2)))=-0.059pH` `=-0.059xx5V` `=-0.295V` `ii.` `pOH=4" ":. pH=14-4=10` `E_((2H^(o+)\|H_(2)))=-0.059pH` `=-0.059xx10V` `=-0.59V`

Discussion

313.	The potential the cell at `25^(@)C` is Given `pK_(b)` of `NH_(4)OH=4.74`A. `0.05V`B. `-0.05V`C. `-0.28V`D. `0.28V`
Answer» Correct Answer - b It is a concentration cell, therefore, `E^(c-)cell=0`. For weak base `NH_(4)OH`, `pOH=(1)/(2)(4.74-log10^(-3))=3.87` `:. pH=14-3.87=10.13` For strong base `NaOH`, `Poh=3,Ph=14-3=11` `:. E_(cell)=-0.059(pH_(c)-pH_(a))` `=-0.059(11-10.13)` `=-0.059xx0.87` `=-0.05V`

Discussion

314.	Given the following cell at `25^(@)C` What will be the potential of the cell ? Given `pK_(a) ` of `CH_(3)COOH=4.74`A. `-0.42V`B. `0.42V`C. `-0.19V`D. `0.19V`
Answer» Correct Answer - a It is a concentration cell, therefore, `E^(c-)._(cell)=0` `pH of W_(A)=(1)/(2)(pK_(a)-log c)` `=(1)/(2)(4.74-log 10^(-3))=3.87` `pH of NaOH=14-3=11` `:. E=-0.059(pH_(c)-pH_(a))` `=-0.059(11-3.87)=-0.42V`

Discussion

315.	A certain quantity of electricity is passed through an aqueous solution of `AgNO_(3)` and cupric salt solution connected in series . The amount of Ag deposited is 1.08 g. then the amount of copper deposited is (At. Mass Cu = 63.5 , Ag = 108) .A. 0.6454 gB. 6.354 gC. 0.3177 gD. 3.177 g
Answer» Correct Answer - C `("Wt. of Cu")/("Wt. of Ag") = ("Eq. of mass of Cu")/("Eq. mass of Ag")` `("Wt. of Cu")/(1.08) = (63.5//2)/(108)` or Wt. of Cu = `0.3177` g

Discussion

316.	The potential of the cell at `25^(@)C ` is Given `pK_(a)` of `CH_(3)COOH` and `pK_(b)` of `NH_(4)OH=4.74`A. `-0.04V`B. `0.04V`C. `-0.189V`D. `0.189V`
Answer» Correct Answer - c It is concentration cell, therefore, `E^(c-)._(cell)=0.` Acidic buffer at anode `:` `:. pH=pK_(a)+log[(Sal t)/(Aci d)]` `=4.74+log .(0.1)/(0.01)=4.74+1=5.74` Basic buffer at cathode `:` `:. pOH=pK_(b)+log[(Sal t)/(Base)]` `=4.74+log.(0.2)/(0.1)` `=4.74+0.03=5.04` `:. pH=14-5.04=8.96` `E_(cell)=-0.059(pH_(c)-pH_(a))` `=-0.050(8.96-5.74)` `=-0.059xx3.22=-0.189V`

Discussion

317.	To deposit `0.6354 g` of copper by electrolysis of aqueous cupric sulpbate solution, the amount of electricity required (in coulmmbs) is.A. 9650B. 4825C. 3860D. 1930
Answer» Correct Answer - D `Cu^(++)+2e^(-)Cu` `E_(Cu)=(63.54)/(2)=31.77` Amount of electricity required to deposit 0.6354gm of Cu `=(96500xx0.6354)/(31.77)=1930` coulomb.

Discussion

318.	Four alkali metals A,B,C and D are having respectively standard electrode potential as -3.05, -1.66, -0.40 and 0.80. Which one will be the most reactive.A. AB. BC. CD. D
Answer» Correct Answer - A More negative is the standard reduction potential, greater is the tendency to lose electrons and hence greater reactivity.

Discussion

319.	A cell, `Ag\|Ag^(+)\|\|Cu^(2+)\|Cu`, initially contains 1 M `Ag^(+)` and `1M" "Cu^(2+)` ions. Calculate the change in cell potential after the passage of 9.65 A of a current of 1 h.
Answer» Quantity of electricity passed passed=`9.65xx60xx60C=34740C` `thereforeAg^(+)` ions deposited`=34740//96500" mole"=0.36` mole `(Ag^(+)+e^(-)toAg)` `Cu^(2+)` ions deposited`=34740//(2xx96500)=0.18` mole `(Cu^(2+)+2e^(-)toCu)` `therefore[Ag^(+)]"left"=1-0.36=0.64M` `[Cu^(2+)]" left"=1-0.18=0.82M` Cell reaction is: `Cu+2Ag^(+)toCu^(2+)+2Ag` `DeltaE=E_(cell)^(@)=(0.0591)/(2)"log"([Cu^(2+)])/([Ag^(+)]^(2))=(0.0591)/(2)"log"(0.82)/((0.64)^(2))=8.89xx10^(-3)V`

Discussion

320.	Which pair of electrolytes could not be distinguished by the products of electrolysis using inert electrodes?A. 1 M `CuSO_(4)` solution, 1 M `CuCl_(2)` solutionB. 1 M KCl solution, 1 M KI solutionC. 1 M `AgNO_(3)` solution, 1 M `Cu(NO_(3))_(2)` solutionD. 1 M KCl solution, 1 M NaCl solution
Answer» Correct Answer - D (a) `CuSO_(4)toCu+O_(2),CuCl_(2)toCu+Cl_(2)` (b) `KCl toH_(2)+Cl_(2),KItoH_(2)+I_(2)` (c) `AgNO_(3)toAg+O_(2),Cu(NO_(3))_(2)toCu+O_(2)` (d) `KCl to H_(2)+Cl_(2),NaCl toH_(2)+Cl_(2)` (e) `CuBrtoCu+Br_(2),CuSO_(4)toCu+O_(2)` Thus, KCl and NaCl solutions give the same products at anode and cathode.

Discussion

321.	Which pair of electrolytes could not be distinguished by the products of electrolysis using inert electrodes ?A. 1 M `CuSO_(4)` solution, 1 M `CuCl_(2)` solutionB. 1 M KCl solution, 1 M KCl solutionC. 1 M `AgNO_(3)` solution, 1 M NaCl solutionD. 1 M KCl solution, 1 M NaCl solution
Answer» Correct Answer - D (d) Electrolysis of both KCl and NaCl solutions liberates `H_(2)` at the cathode and `Cl_(2)` at the anode. Thus, the distinction cannot be made by the gases evolved at the two electrodes. However, resulting solutions will be different. KOH in case of KCl and NaOH in case of NaCl.

Discussion

322.	Which pair of electrolytes could not be distinguished by the products of electrolysis using inert electrodes.A. 1M `CuSO_(4)` solution, `1M CuCl_(2)` solutionB. 1M KCl solution, 1M KI solutionC. 1 M `AgNO_(3)` solution `1M Cu(NO_(3))` solutionD. 1M `CuBr_(2)` solution `1M CuSO_(4)` solution.
Answer» Correct Answer - A (A) `CuSO_(4)rarrCu+O_(2):CuCl_(2)rarrCu+I_(2)` (B) `KClrarrH_(2)+Cl_(2):KIrarrH_(2)+I_(2)` (C) `AgNO_(3)rarrAG+O_(2),:NaClrarrH_(2)+Cl_(2)` (D) `KClrarrH_(2)+Cl_(2):NaClrarrH_(2)+Cl_(2)` (E) `CuBrrarrCu+Br_(2):CuSO_(4)rarrCu+O_(2)` Thus KCl and NaCl solution give the same products at anode and cathode.

Discussion

323.	The electrode potentials for `Cu^(2+)+e^(-)toCu^(+)` and `Cu^(+)e^(-)toCu_(s)` are `+0.15`V and `+0.50V` respectively the value of `E_((Cu^(2+))/(Cu))^(@)` will be?A. 0.150VB. 0.500VC. 0.325VD. 0.650V
Answer» Correct Answer - C `Cu^(2+)+erarrCu^(+)` `DeltaG_(1)=-nFE^(@)=-1xxFxx0.15` `Cu^(+)+erarrCu,DeltaG_(2)=-1xxFxx0.50` `Cu^(2+)+2erarrCu` `DeltaG_(3)=-1xxFxxE_(Cu^(2+)//Cu)^(@)` `DeltaG_(3)=DeltaG_(1)+DeltaG_(2)` `-2FE_(Cu^(2+)//Cu)^(@)= -0.15F+(-0.50E)` or `-2E_(Cu^(2+)//Cu)^(@)=-0.15-0.50=-0.65` or `E_(Cu^(2+)//Cu)^(@)=(0.65)/(2)=0.325V`.

Discussion

324.	Standard reduction potential of the half reaction are given below `F_(2)(g)+2e^(-)to2F^(-)(aq)," "E^(@)=+2.85V` `Cl_(2)(g)+2e^(-)to2Cl^(-)(aq)," "E^(@)=+1.36V` `Br_(2)(l)+2e^(-)to2Br^(-)(aq)," "E^(@)=+1.06V` `I_(2)(s)+2e^(-)to2I^(-)(aq)," "E^(@)=+0.53V` the strongest oxidising and reducig agents respectively areA. `F_(2) and I^(-)`B. `Br_(2) and Cl^(-)`C. `Cl_(2) and Br^(-)`D. `Cl_(2) and I_(2)`
Answer» Correct Answer - A `E^(o)` more positive, reducing agent will be greater.

Discussion

325.	Limiting molar conductivity of `NH_4 OH` I e., `Lambda_m (NH_4 OH)` is equal to .A. `Lambda_m (NH_4 OH) + Lambda_m(NH_4Cl)-Lambda_m(HCl)`B. `Lambda_m (NH_4 OH) + Lambda_m(Na_OH)-Lambda_m(HCl)`C. `Lambda_m (NH_4 OH) + Lambda_m(NsCl)-Lambda_m(HCl)`D. `Lambda_m (NH_4 OH) + Lambda_m(NaCl)-Lambda_m(HCl)`
Answer» Correct Answer - B ` Lambda_m^@ (NH_4 Cl) = Lambda_m^@ (NH_4^+) + Lambda_m^@ (Cl^(-) )` …(i) ` A_m^@ (NAOH) = Lambda_m^@(Na^+) + Lambda_m^2 (OH^(-))` ..(ii) `Lambda_m^2 (NaCl)= Lambda_m^@ (Na^+) +Lambda_m^@ (Cl^(-))` …(iii) By (i) + (ii) + (iii) `Lambda_m^@ (NH_4 Cl) =Lambda_m^2 (NH_4^+) + [email protected] (Cl^(-))` `=Lambda_m^@ (NH_4Cl) = Lambda) _m^@ (NaOH)` ` = Lambda_m^2 (NACl)`.

Discussion

326.	`E^(@)` for `Cl_2`(g) `+2e^(-)to2Cl^-`(aq) is 1.36 V, `E^(@)` for `Cl^-` (aq) `to1//2cl_2`(g)`+e^-` is:A. 1.36VB. -1.36VC. -0.68D. 0.68V
Answer» Correct Answer - B

Discussion

327.	The electrode potentials for `Cu^(2+)+e^(-)toCu^(+)` and `Cu^(+)e^(-)toCu_(s)` are `+0.15`V and `+0.50V` respectively the value of `E_((Cu^(2+))/(Cu))^(@)` will be?A. 0.650VB. 0.150VC. 0.500VD. 0.325V
Answer» Correct Answer - D `Cu^(2+)(aq)+ e^(-)rarrCu^(+)(aq):E_(1)^(@)=0.15V` `DeltaG_(1)^(@)=-nE_(2)^(@)F=-0.15F` `Cu^(+)(aq)+e^(-)rarrCu(s):E_(2)^(@)=0.50V` `DeltaG_(2)^(@)=-nE_(2)^(@)F=-0.50F` `Cu^(2+)(aq)+2e^(-)rarrCu(s):E_(3)^(@)= ?` `DeltaG_(3)^(@)=-nE_(3)^(@)F= -2E_(3)^(@)F` `DeltaG_(3)^(@)=DeltaG_(1)^(@)+DeltaG_(2)^(@)` `-2E_(3)^(@)F=-0.15F=0.50F` `therefore E_(3)^(@)=(0.65)/(2)=0.325V`

Discussion

328.	Standard electrode potential of three metal `X, Y` and `Z` are `-1.2 V, +0.5 V` and `-3.0 V` respectively. The reducing power of these metals will be:A. `Xgt Ygt Z`B. `Ygt Zgt X`C. `Ygt Xgt Z`D. `Zgt Xgt Y`
Answer» Correct Answer - D A reducing agent reduces the other and gets oxidized. Thus, the reducing power is inversely related to the value of reducation potential i.e., higher the reduction potential, lower the reducing power. Thus, the correct order is `Z(-3.0 V)gt X(-1.2 V)gt Y(+0.5 V)`

Discussion

329.	Standrd reduction potentials of the half reactions are given below : `F_2 (g) +re^- rarr 2 F^(-) (aq) E^@ = + 2.85 V` ` Cl_2 (g) +2e^- rarr 2 Cl^-(aq) , E^2 = + 1.36 V` `Br _2 (i) + 2 e^- rarr 2Br (aq) , E^2 = + 1. 06 V` `I_2 (s) + 2 e^- rarr 2I^(-) (aq) , E^2 = + . 53 V`. The strongest oxidizing and reducing agents respectivelyA. ` F_2` and `I^(-)`B. ` Br_(2)` and `C l^(-)`C. `Cl_2` and `Br^(-)`D. ` Cl_2` and `I_2`
Answer» Correct Answer - A The given values are `E_(RP)^@` More si `E_(RP)^@` , more is the tendency to gin electron or to show reduction or to show strong oxedant nature .

Discussion

330.	The reduction potential of hydrogen half cell will be negative if :A. `p(H_(2))=1 "atm and" [H^(+)]=1.0M`B. `p(H_(2))=2 "atm and"[H^(+)]=1.0M`.C. `p(H_(2))=2"atm and"[H^(+)]=2.0M`D. `p(H_(2))=1 "atm and" [H^(+)]=2.0M`
Answer» Correct Answer - B `2H^(+)+2e^(-)rarrH_(2)` `E_(2H^(+)//H_(2))=E_(2H^(+)//H_(2))^(@)-(0.059)/(2)"log"(p(H_(2)))/([H^(+)]^(2))` `E_(2H^(+)//H_(2))` will be negative if `p(H_(2))gt[H^(+)]^(2)` i.e., `p(H_(2))=2` atm and `[H^(+)]=1`M.

Discussion

331.	Standard electrode potential of three metal `X, Y` and `Z` are `-1.2 V, +0.5 V` and `-3.0 V` respectively. The reducing power of these metals will be:A. `ZgtXgtY`B. `XgtYgtZ`C. `YgtZgtX`D. `YgtXgtZ`.
Answer» Correct Answer - A R.P . Of X=-1.2V R.P. of Y=+0.4Y R.P. of Z=-3.0V More the R.P. easier to reduce i.e. it is a stronger O.A. or weaker R.A. ltbr. Thus R.P `prop(1)/("reducing power")` Thus correct order of reduc ing power is: `ZgtXgtY`

Discussion

332.	If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the correct relationships for the values of `DeltaG^(ɵ)` and `K_(eq)`?A. ` DeltaG^@ gt 0 , K(eq) lt 1`B. ` Delta G^@ gt 0K_(eq) , K)(eq) gt 1`C. ` Delta G^@ lt 0, K_(eq) gt 1`D. ` Zgt X gt Y`
Answer» Correct Answer - A `E^@ = (0. 059)/n log K_(eq) and Delta G^@ More si E_(CP)^@`F `:. Delta G^@ + ve , E^@` will be -and`K_(eq) lt` one shoule not writed Delta `G^@ gt 0`.

Discussion

333.	If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the correct relationships for the values of `DeltaG^(ɵ)` and `K_(eq)`?A. `DeltaG^(@) gt 0,K_(eq) lt 1`B. `DeltaG^(@) gt 0,K_(eq) gt 1`C. `DeltaG^(@) lt 0,K_(eq) gt 1`D. `DeltaG^(@) lt 0,K_(eq) lt 1`
Answer» Correct Answer - A (a) Given, `E_(cell)^(@)=-ve` The relation between `DeltaG^(@) and E_(cell)^(@)` is given as `DeltaG^(@)=-nFE_(cell)^(@)`…(i) If `E_(cell)^(@)` is negative, so `DeltaG^(@)` comes out to be positive. Again, relation between `DeltaG^(@)` and `K_(eq)` is given as `DeltaG^(@) =- 2.303 nRT log K_(eq)` ...(ii) From Eq. (i) we get that `DeltaG^(@)` is positive . Now , if `DeltaG^(@)` is positive then `K_(eq)` comes out to be negative from eq (ii). i.e `DeltaG^(@) gt 1 and K_(eq) lt 1` Short trick As `E_(cell)^(@)` is negative so reaction is non-spontaneous or you can say reaction is moving in backward direction. For non-spontaneous reaction. `DeltaG^(@)` is always positive and `K_(eq)` is always less than 1.

Discussion

334.	If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the correct relationships for the values of `DeltaG^(ɵ)` and `K_(eq)`?A. `DeltaG^(@)lt0,k_(eq)lt1`B. `DeltaG^(@)gt0,k_(eq)lt1`C. `DeltaG^(@)gt0,k_(eq)gt1`D. `DeltaG^(@)lt0,k_(eq)gt1`.
Answer» Correct Answer - B `DeltaG^(@)=-nE^(@)F` as `E^(@)lt0therefore k_(Eq)lt1`.

Discussion

335.	`E_(Cell)^(Theta)=1.1V` for Daniell cell. Which of the following expressions are correct description of state of equilibrium in this cell?A. `1.1=K_(c)`B. `(2.303RT)/(2F)logK_(c)=1.1`C. `logK_(c)=(2.2)/(0.059)`D. `logK_(c)=1.1`
Answer» Correct Answer - B::C For Daniell cell `Zn+Cu^(2+)toZn^(2+)+Cu,E_(cell)=E_(cell)^(@)-(2.303RT)/(nF)"log"([Zn^(2+)])/([Cu^(2+)])` At equilibrium `E_(cell)=0`. Hence, `E_(cell)^(@)=(2.303RT)/(2F)logK_(c)=1.1` (given) i.e.., (b) Putting `(2.303RT)/(F)=0.059` at 298K, `(0.059)/(2)logK_(c)=1.1` or log`K_(c)=(2.2)/(0.059)` i.e., (c).

Discussion

336.	Assertion When acidic solution of `KMnO_(4)` is added from burette to a beaker containing oxalic acid solution, following reaction occur: `MnO_(4)^(-)+H_(2)C_(2)O_(4)+H^(+)rarrMn^(2+)+CO_(2)` However the experimentalist does not experiences any electric shock. Reason The above redox reaction is non-spontaneous and no electricity is produced during the reaction.A. Both assertion and reason are correct and reason is the correct but reason is the correct explanation of the assertion.B. Both assertion and reason are correct but reason is not the correct explantion of assertion.C. Assertion is correct but reason is wrong.D. Assertion is wrong but reason is correct.
Answer» Correct Answer - c

Discussion

337.	Class 12 Chemistry MCQ Questions of Electrochemistry with Answers?
Answer» MCQ Questions for Class 12 Chapter Wise with Answers were Prepared based on Latest Exam Pattern and syllabus. Students can solve Class 12 MCQ Questions with Answers to know their preparation level. Electrochemistry is worried about charges with their development and movement starting with one medium then onto the next. An ultimate unit of charge is that conveyed by the electron, electrons are significant in electrochemistry and their capacities here are like a portion of those which they display in related teaches all the more normally viewed as the area of physics. Important Class 12 Chemistry MCQ Questions of Electrochemistry with Answers are available here by Sarthaks eConnect. It helps students understand the concept very well. Refer to more Chapter-wise MCQ Questions for Class 12 Chemistry Electrochemistry for all subjects. Practice MCQ Question for Class 12 Chemistry chapter-wise 1. If 96500 coulomb electricity is passed through CuSO₄solution, it will liberate (a) 63.5 gm of Cu (b) 31.76 gm of Cu (c) 96500 gm of Cu (d) 100 gm of Cu 2. Fused NaCl on electrolysis gives ………….. on cathode. (a) Chlroine (b) Sodium (c) Sodium amalgam (d) Hydrogen 3. Which of the following is a secondary cell? (a) Leclanche cell (b) Lead storage battery (c) Concentration cell (d) All of these 4. Faraday’s law of electrolysis is related to (a) Atomic number of cation (b) Speed of cation (c) Speed of anion (d) Equivalent weight of electrolyte 5. How much electricity in terms of Faraday is required to produce 100 g of Ca from molten CaCl₂? (a) 1F (b) 2F (c) 3F (d) 5F 6. How many coulombs of electricity is required to reduce 1 mole of \(Cr_2O_7^{2-}\)in acidic medium? (a) 4 \(\times\) 96500 C (b) 6 \(\times\) 96500 C (c) 2 \(\times\) 96500 C (d) 1 \(\times\) 96500 C 7. Effect of dilution on conductivity of solution: (a) Increases (b) Decreases (c) Unchanged (d) None of the above 8. What flows in the internal circuit of a galvanic cell? (a) Ions (b) Electrons (c) Electricity (d) Atoms 9. Which of the following statements about galvanic cell is incorrect (a) anode is positive (b) oxidation occurs at the electrode with lower reduction potential (c) cathode is positive (d) reduction occurs at cathode 10. The electric charge for electrode decomposition of one gram equivalent of a substance is (a) one ampere per second (b) 96500 coulombs per second (c) one ampere for one hour (d) charge on one mole of electrons 11. In electrolysis of dilute H₂SO₄ using platinum electrodes (a) H₂ is evolved at cathode (b) NH₂ is produced at anode (c) Cl₂ is obtained at cathode (d) O₂ is produced 12. The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called............. (a) Cell potentials (b) Cell emf (c) Potential difference (d) Cell voltage 13. The reference electrode is made by using (a) ZnCl₂ (b) CuSO₄ (c) HgCl₂ (d) Hg₂Cl₂ 14. The standard hydrogen electrode potential is zero, because (a) hydrogen oxidized easily (b) electrode potential is considered as zero (c) hydrogen atom has only one electron (d) hydrogen is a very light element 15. Without losing its concentration ZnCl₂ solution cannot be kept in contact with (a) Au (b) Al (c) Pb (d) Ag 16. Standard electrode potential of three metals X, Y and Z are – 1.2 V, + 0.5 V and – 3.0 V, respectively. The reducing power of these metals will be : (a) Y > Z > X (b) X > Y > Z (c) Z > X > Y (d) X > Y > Z 17. Standard electrode potential for Sn⁴⁺/ Sn²⁺ couple is + 0.15 V and that for the Cr³⁺/ Cr couple is – 0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be (a) + 1.19 V (b) + 0.89 V (c) + 0.18 V (d) + 1.83 V 18. The oxidation potentials of A and B are +2.37 and +1.66 V respectively. In chemical reactions (a) A will be replaced by B (b) A will replace B (c) A will not replace B (d) A and B will not replace each other 19. Which of the following statement is not correct about an inert electrode in a cell? (a) It does not participate in the cell reaction (b) It provides surface either for oxidation or for reduction reaction (c) It provides surface for conduction of electrons (d) It provides surface for redox reaction 20. Which of the following is the use of electrolysis? (a) Electrorefining (b) Electroplating (c) Both (a) & (b) (d) None of these 21. The amount of electricity that can deposit 108 g of Ag from AgNO₃ solution is: (a) 1 F (b) 2 A (c) 1 C (d) 1 A 22. The cell constant of a conductivity cell.................. (a) changes with change of electrolyte (b) changes with change of concentration of electrolyte (c) changes with temperature of electrolyte (d) remains constant for a cell 23. The standard reduction potential at 298K for the following half cells are given: Which is the strongest reducing agent? (a) Zn(s) (b) Cr(s) (c) Both (d) None of these 24. In electroplating the article to be electroplated is made (a) cathode (b) anode (c) either cathode or anode (d) simply suspended in the electrolytic bath 25. A device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy is known as : (a) Fuel cell (b) Electrolytic cell (c) Dynamo (d) Ni -Cd cell Answer : 1. Answer : (b) 31.76 gm of Cu Explanation: At the cathode \(Cu^{2+}(aq)+2e^-\longrightarrow\)\(Cu(s)\) At the anode \(4OH^-(aq)\longrightarrow\) \(2H_2O(l)+O_2(g)+4e^-\) Faraday’s constant = 96500C / mol To deposit 1 mole of copper, we need \(2\times96500\) So, 96500C will deposit 0.5 moles of copper \(=0.\times63.5=31.75g\) 2. Answer : (b) Sodium Explanation: Fused on electrolysis gives on cathode gives sodium. 3. Answer : (b) Lead storage battery Explanation: Lead storage cells and nickel-cadmium cells are examples of secondary cells. 4. Answer : (d) Equivalent weight of electrolyte Explanation: Equivalent weight is the ratio of the atomic mass of metal and the number of electrons that are required for the reduction of the cation. Hence, Faraday's law of electrolysis is related to the equivalent mass of electrolyte. 5. Answer : (d) 5F Explanation: \(CaCl_2\longrightarrow\) \(Ca^{2+}+2Cl^-\) Ca is undergoing oxidation i.e, \(Ca^{2+}+2e^-\longrightarrow\) \(Ca\) We can observe that 40g of Ca takes \(2e^-\) charge, Now, 100g of Ca = 5F 6. Answer : (b) 6 \(\times\) 96500 C Explanation: In acidic medium dichromate ions undergoes reduction and the reaction is, \(Cr_2O_7+14H^++6e^-\longrightarrow\) \(2Cr^{3+}+7H_2O\) So, 1 mole of \(Cr_2O_7^{2-}\) ion requires 6 moles of electrons i.e. 6 \(\times\) 96500 C of electricity. 7. Answer : (a) Increases Explanation: When a solution is diluted, there are more ions and they have more space to move, i.e. ions are more far away from each other and mobility of ions increase which leads to increase in molar conductivity of the solution. Hence when a solution is diluted, conductivity decreases and molar conductivity increases. 8. Answer : (a) Ions Explanation: The galvanic cell's two fundamental components will be these two electrodes. The reduction half-reaction should happen at the cathode, whereas the oxidation half-reaction should happen at the anode. Therefore, ions flow in the inner circuit or the internal circuit of the galvanic cell. 9. Answer : (a) anode is positive Explanation: Anode has negative polarity in galvanic cell. 10. Answer : (d) charge on one mole of electrons Explanation: Charge on one mole of electrons is equal to one Faraday, so one gram equivalent of a substance will be deposited by one mole of electrons. 1 g equivalent of a substance means one equivalent of substance which have total charge of one mole of electron. 11. Answer : (a) H₂ is evolved at cathode Explanation: H₂ is liberated at Cathode while O₂ is liberated at anode. 12. Answer : (b) Cell emf Explanation: Cell emf is the energy provided by a cell or battery per coulomb of charge passing through it.So, the difference between the electrode potentials of two electrodes when no current is drawn through the cell is called cell emf. 13. Answer : (d) Hg₂Cl₂ Explanation: The most commonly used reference electrode is calomel electrode. Its reduction potential depends upon the concentration of KCl solution used. The calomel electrode is made by using Hg₂Cl₂. 14. Answer : (b) electrode potential is considered as zero Explanation: In electrochemistry, hydrogen is taken to be the reference to measure the potential and hence to form a basis for comparison with all other electrode reactions, hydrogen's standard electrode potential is declared to be zero volts at all temperatures. 15. Answer : (b) Al Explanation: Only Al(E^∘=−1.66V) has lower electrode potential than Zn(E^∘ =−0.76V) and hence it will reduce Zn ²⁺ to Zn while all others have higher electrode potentials than Zn. 16. Answer : (d) X > Y > Z Explanation: Higher the reduction potential lesser is reducing power and vice-versa. Hence, the order of reducing power is Z > X > Y. 17. Answer : (b) + 0.89 V Explanation: E^∘_cell = E^∘_{cathode (RP)} − E^∘_{anode (RP)} = 0.15 − (−0.74) = + 0.89 V 18. Answer : (b) A will replace B Explanation: A metal with higher O.P. (i.e. A) replaces a metal with lower O.P. (i.e. B) from its aqueous solution. 19. Answer : (d) It provides surface for redox reaction Explanation: The inert electrode is an electrode that serves only as a source or sinks for electrons. Precious metals, mercury, and carbon are typically used as inert electrodes. It provides a surface for oxidation or reduction reaction but not for the redox reaction. 20. Answer : (d) None of these Explanation: The main application of electrolysis are: Electrorefining, Electroplating, Extraction of metals. 21. Answer : (a) 1 F Explanation: 1 Faraday of charge discharges E gram of an ion, where E is the equivalent weight. For silver, the value of E is 108 g. Thus, the amount of electricity that can deposit 108 g of silver form AgNO₃solution is one Faraday. 22. Answer : (d) remains constant for a cell Explanation: The cell constant, K, is equal to the distance in cm between the probe's electrodes divided by the surface area of the electrodes in cm² .For solutions with low conductivities the electrodes can be placed closer together or made larger so that the cell constant is less than one. 23. Answer : (a) Zn(s) Explanation: Zinc is the strongest reducing agent in the following above given reactions because the standard reduction potentials of the Zinc, which is given, is the most negative than others. 24. Answer : (a) cathode Explanation: The material to be electroplated, should be made cathode. The positive radical from the electrolyte goes to the cathode and gets deposited there by reduction process. 25. Answer : (a) Fuel cell Explanation: A fuel cell is a device that converts the energy of combustion of fuels like hydrogen and methane, directly into electrical energy. Electrolytic cell converts electrical energy into chemical energy. Click here for MCQ Question for Electrochemistry Class 12

337.

Class 12 Chemistry MCQ Questions of Electrochemistry with Answers?

Answer»

MCQ Questions for Class 12 Chapter Wise with Answers were Prepared based on Latest Exam Pattern and syllabus. Students can solve Class 12 MCQ Questions with Answers to know their preparation level. Electrochemistry is worried about charges with their development and movement starting with one medium then onto the next. An ultimate unit of charge is that conveyed by the electron, electrons are significant in electrochemistry and their capacities here are like a portion of those which they display in related teaches all the more normally viewed as the area of physics.

Important Class 12 Chemistry MCQ Questions of Electrochemistry with Answers are available here by Sarthaks eConnect. It helps students understand the concept very well. Refer to more Chapter-wise MCQ Questions for Class 12 Chemistry Electrochemistry for all subjects.

Practice MCQ Question for Class 12 Chemistry chapter-wise

1. If 96500 coulomb electricity is passed through CuSO₄solution, it will liberate

(a) 63.5 gm of Cu
(b) 31.76 gm of Cu
(c) 96500 gm of Cu
(d) 100 gm of Cu

2. Fused NaCl on electrolysis gives ………….. on cathode.

(a) Chlroine
(b) Sodium
(c) Sodium amalgam
(d) Hydrogen

3. Which of the following is a secondary cell?

(a) Leclanche cell
(b) Lead storage battery
(c) Concentration cell
(d) All of these

4. Faraday’s law of electrolysis is related to

(a) Atomic number of cation
(b) Speed of cation
(c) Speed of anion
(d) Equivalent weight of electrolyte

5. How much electricity in terms of Faraday is required to produce 100 g of Ca from molten CaCl₂?

(a) 1F
(b) 2F
(c) 3F
(d) 5F

6. How many coulombs of electricity is required to reduce 1 mole of \(Cr_2O_7^{2-}\)in acidic medium?

(a) 4 \(\times\) 96500 C
(b) 6 \(\times\) 96500 C
(c) 2 \(\times\) 96500 C
(d) 1 \(\times\) 96500 C

7. Effect of dilution on conductivity of solution:

(a) Increases
(b) Decreases
(c) Unchanged
(d) None of the above

8. What flows in the internal circuit of a galvanic cell?

(a) Ions
(b) Electrons
(c) Electricity
(d) Atoms

9. Which of the following statements about galvanic cell is incorrect

(a) anode is positive
(b) oxidation occurs at the electrode with lower reduction potential
(c) cathode is positive
(d) reduction occurs at cathode

10. The electric charge for electrode decomposition of one gram equivalent of a substance is

(a) one ampere per second
(b) 96500 coulombs per second
(c) one ampere for one hour
(d) charge on one mole of electrons

11. In electrolysis of dilute H₂SO₄ using platinum electrodes

(a) H₂ is evolved at cathode
(b) NH₂ is produced at anode
(c) Cl₂ is obtained at cathode
(d) O₂ is produced

12. The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called.............

(a) Cell potentials
(b) Cell emf
(c) Potential difference
(d) Cell voltage

13. The reference electrode is made by using

(a) ZnCl₂
(b) CuSO₄
(c) HgCl₂
(d) Hg₂Cl₂

14. The standard hydrogen electrode potential is zero, because

(a) hydrogen oxidized easily
(b) electrode potential is considered as zero
(c) hydrogen atom has only one electron
(d) hydrogen is a very light element

15. Without losing its concentration ZnCl₂ solution cannot be kept in contact with

(a) Au
(b) Al
(c) Pb
(d) Ag

16. Standard electrode potential of three metals X, Y and Z are – 1.2 V, + 0.5 V and – 3.0 V, respectively. The reducing power of these metals will be :

(a) Y > Z > X
(b) X > Y > Z
(c) Z > X > Y
(d) X > Y > Z

17. Standard electrode potential for Sn⁴⁺/ Sn²⁺ couple is + 0.15 V and that for the Cr³⁺/ Cr couple is – 0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be

(a) + 1.19 V
(b) + 0.89 V
(c) + 0.18 V
(d) + 1.83 V

18. The oxidation potentials of A and B are +2.37 and +1.66 V respectively. In chemical reactions

(a) A will be replaced by B
(b) A will replace B
(c) A will not replace B
(d) A and B will not replace each other

19. Which of the following statement is not correct about an inert electrode in a cell?

(a) It does not participate in the cell reaction
(b) It provides surface either for oxidation or for reduction reaction
(c) It provides surface for conduction of electrons
(d) It provides surface for redox reaction

20. Which of the following is the use of electrolysis?

(a) Electrorefining
(b) Electroplating
(c) Both (a) & (b)
(d) None of these

21. The amount of electricity that can deposit 108 g of Ag from AgNO₃ solution is:

(a) 1 F
(b) 2 A
(c) 1 C
(d) 1 A

22. The cell constant of a conductivity cell..................

(a) changes with change of electrolyte
(b) changes with change of concentration of electrolyte
(c) changes with temperature of electrolyte
(d) remains constant for a cell

23. The standard reduction potential at 298K for the following half cells are given: Which is the strongest reducing agent?

(a) Zn(s)
(b) Cr(s)
(c) Both
(d) None of these

24. In electroplating the article to be electroplated is made

(a) cathode
(b) anode
(c) either cathode or anode
(d) simply suspended in the electrolytic bath

25. A device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy is known as :

(a) Fuel cell
(b) Electrolytic cell
(c) Dynamo
(d) Ni -Cd cell

Answer :

1. Answer : (b) 31.76 gm of Cu

Explanation:

At the cathode

\(Cu^{2+}(aq)+2e^-\longrightarrow\)\(Cu(s)\)

At the anode

\(4OH^-(aq)\longrightarrow\) \(2H_2O(l)+O_2(g)+4e^-\)

Faraday’s constant = 96500C / mol

To deposit 1 mole of copper, we need \(2\times96500\) So, 96500C will deposit 0.5 moles of copper

\(=0.\times63.5=31.75g\)

2. Answer : (b) Sodium

Explanation: Fused on electrolysis gives on cathode gives sodium.

3. Answer : (b) Lead storage battery

Explanation: Lead storage cells and nickel-cadmium cells are examples of secondary cells.

4. Answer : (d) Equivalent weight of electrolyte

Explanation: Equivalent weight is the ratio of the atomic mass of metal and the number of electrons that are required for the reduction of the cation. Hence, Faraday's law of electrolysis is related to the equivalent mass of electrolyte.

5. Answer : (d) 5F

Explanation: \(CaCl_2\longrightarrow\) \(Ca^{2+}+2Cl^-\)

Ca is undergoing oxidation i.e,

\(Ca^{2+}+2e^-\longrightarrow\) \(Ca\)

We can observe that 40g of Ca takes \(2e^-\) charge,

Now, 100g of Ca = 5F

6. Answer : (b) 6 \(\times\) 96500 C

Explanation: In acidic medium dichromate ions undergoes reduction and the reaction is,

\(Cr_2O_7+14H^++6e^-\longrightarrow\) \(2Cr^{3+}+7H_2O\)

So, 1 mole of \(Cr_2O_7^{2-}\) ion requires 6 moles of electrons i.e. 6 \(\times\) 96500 C of electricity.

7. Answer : (a) Increases

Explanation: When a solution is diluted, there are more ions and they have more space to move, i.e. ions are more far away from each other and mobility of ions increase which leads to increase in molar conductivity of the solution. Hence when a solution is diluted, conductivity decreases and molar conductivity increases.

8. Answer : (a) Ions

Explanation: The galvanic cell's two fundamental components will be these two electrodes. The reduction half-reaction should happen at the cathode, whereas the oxidation half-reaction should happen at the anode. Therefore, ions flow in the inner circuit or the internal circuit of the galvanic cell.

9. Answer : (a) anode is positive

Explanation: Anode has negative polarity in galvanic cell.

10. Answer : (d) charge on one mole of electrons

Explanation: Charge on one mole of electrons is equal to one Faraday, so one gram equivalent of a substance will be deposited by one mole of electrons. 1 g equivalent of a substance means one equivalent of substance which have total charge of one mole of electron.

11. Answer : (a) H₂ is evolved at cathode

Explanation: H₂ is liberated at Cathode while O₂ is liberated at anode.

12. Answer : (b) Cell emf

Explanation: Cell emf is the energy provided by a cell or battery per coulomb of charge passing through it.So, the difference between the electrode potentials of two electrodes when no current is drawn through the cell is called cell emf.

13. Answer : (d) Hg₂Cl₂

Explanation: The most commonly used reference electrode is calomel electrode. Its reduction potential depends upon the concentration of KCl solution used. The calomel electrode is made by using Hg₂Cl₂.

14. Answer : (b) electrode potential is considered as zero

Explanation: In electrochemistry, hydrogen is taken to be the reference to measure the potential and hence to form a basis for comparison with all other electrode reactions, hydrogen's standard electrode potential is declared to be zero volts at all temperatures.

15. Answer : (b) Al

Explanation: Only Al(E^∘=−1.66V) has lower electrode potential than Zn(E^∘ =−0.76V) and hence it will reduce Zn ²⁺ to Zn while all others have higher electrode potentials than Zn.

16. Answer : (d) X > Y > Z

Explanation: Higher the reduction potential lesser is reducing power and vice-versa. Hence, the order of reducing power is Z > X > Y.

17. Answer : (b) + 0.89 V

Explanation: E^∘_cell = E^∘_{cathode (RP)} − E^∘_{anode (RP)}

= 0.15 − (−0.74)

= + 0.89 V

18. Answer : (b) A will replace B

Explanation: A metal with higher O.P. (i.e. A) replaces a metal with lower O.P. (i.e. B) from its aqueous solution.

19. Answer : (d) It provides surface for redox reaction

Explanation: The inert electrode is an electrode that serves only as a source or sinks for electrons. Precious metals, mercury, and carbon are typically used as inert electrodes. It provides a surface for oxidation or reduction reaction but not for the redox reaction.

20. Answer : (d) None of these

Explanation: The main application of electrolysis are: Electrorefining, Electroplating, Extraction of metals.

21. Answer : (a) 1 F

Explanation: 1 Faraday of charge discharges E gram of an ion, where E is the equivalent weight. For silver, the value of E is 108 g. Thus, the amount of electricity that can deposit 108 g of silver form AgNO₃solution is one Faraday.

22. Answer : (d) remains constant for a cell

Explanation: The cell constant, K, is equal to the distance in cm between the probe's electrodes divided by the surface area of the electrodes in cm² .For solutions with low conductivities the electrodes can be placed closer together or made larger so that the cell constant is less than one.

23. Answer : (a) Zn(s)

Explanation: Zinc is the strongest reducing agent in the following above given reactions because the standard reduction potentials of the Zinc, which is given, is the most negative than others.

24. Answer : (a) cathode

Explanation: The material to be electroplated, should be made cathode. The positive radical from the electrolyte goes to the cathode and gets deposited there by reduction process.

25. Answer : (a) Fuel cell

Explanation: A fuel cell is a device that converts the energy of combustion of fuels like hydrogen and methane, directly into electrical energy. Electrolytic cell converts electrical energy into chemical energy.

Click here for MCQ Question for Electrochemistry Class 12

Discussion

338.	If standard reduction potentials for Pb, K, Zn and Cu are -0.126 V, -2.925 V, -0.763 V and 0.337 V, the decreasing order of reducing power is :(a) Zn > Pb > K > Cu (b) Cu > Pb > Zn > K (c) K > Zn > Pb > Cu (d) K > Pb > Cu > Zn
Answer» Option : (c) K > Zn > Pb > Cu

Discussion

339.	The standard reduction potential of Pb and Zn electrodes are `-0.126` and `-0.763 V` respectively. The cell equation will beA. `Pb^(2+)+Zn rarr Pb+Zn^(2+)`B. `Pb^(4+)+2 Zn rarr Pb+2 Zn^(2+)`C. `Zn^(2+)+Pb rarr Zn+Pb^(2+)`D. None of the above
Answer» Correct Answer - C The `E_(red)^(@)` of `Zn` is lower than that of `Pb`, hence `Zn` is a better reducing agent than `Pb`. `:.` The correct cell reaction will be `Zn^(2+)+Pb rarr Zn+Pb^(2+)`

Discussion

340.	Given that, `Ni^(2+)//Ni=0.25 V, Cu^(2+)//Cu=0.34 V`, `Ag^(+)//Ag=0.80 V` and `Zn^(2+)//Zn=-0.76 V` Which of the following reaction under standard condition will not take place in the specified direction ?A. `Ni^(2+) (aq.)+Cu(s) rarr Ni(s) +Cu^(2+) (aq.)`B. `Cu(s)+2Ag^(+)(aq.) rarr Cu^(2+) (aq.)+2Ag(s)`C. `Cu(s)+2H^(+) (aq.) rarr Cu^(2+) (aq.)+H_(2)(g)`D. `Zn(s)+2H^(+)(aq.) rarr Zn^(2+) (aq.)+3H_(2)(g)`
Answer» Correct Answer - A::C::D

Discussion

341.	The `E^(@)` value of Zn is -0.76 V while that of Cu is +0.34 V. Do these values help in locating the relative positions of the electrodes in the electrochemical series ?
Answer» Correct Answer - Yes, they do help.

Discussion

342.	What iis electrochemical series ? How does it help in predicting whether a particular redox reaction is feasible in a given direaction or not.
Answer» Redox reaction is feasible if e.m.f. of cell is positive.

Discussion

343.	The cell constant of a conductivity cellA. change with change of electrolyteB. changes with change of concentration of electrolyteC. changes with temperture of electrolyteD. remains constant for a cell
Answer» Correct Answer - D

Discussion

344.	Assertion:A standard hydrogen electrode is also called reversible electrode. Reason : Standard hydrogen electrode can act both as anode as well as cathode in an electrochemical cell.A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false .
Answer» Correct Answer - A

Discussion

345.	Assertion:Electrolytic cell uses electrical energy to carry non-spontaneous chemical reactions. Reason : Chemical energy of a spontaneous redox reaction can be converted into electrical energy.A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false .
Answer» Correct Answer - B

Discussion

346.	Assertion: an electrochemical cell can be set up only if the redox reaction is spontaneous. Reason: A reaction is spontaneous if free energy change is negative.A. if both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If the assertion and reason both are false.
Answer» Correct Answer - B If redox reaction is spontaneous, `DeltaG` is -ve and hence `E^(@)` is positive. `-DeltaG^(@)=nFE^(@)` cell.

Discussion

347.	The standard reduction potential of `Cu^(2+)//Cu` and `Cu^(2+)//Cu^(+)` are `0.339 V` and `0.153 V` respectively. The standard electrode potential of `Cu^(+)//Cu` half cell is :A. `0.525 V`B. `0.827 V`C. `0.184 V`D. `0.490 V`
Answer» Correct Answer - A `{:(Cu^(2+)+2erarrCu,-DeltaG_(1)^(@)=n_(1)E_(1)^(@)F),(ul(Cu^(2+)+erarrCu^(+),-DeltaG_(2)^(@)=n_(2)E_(2)^(@)F)):}` Half reaction `Cu^(+) + e rarrCu`, `-DeltaG_(3)^(@) = -Delta_(1)^(@) + Delta_(2)^(@) = n_(3)E_(3)^(@)F` or `n xx E_(3)^(@)F = n_(1)E_(1)^(@)F - n_(2)E_(2)^(@)F` `E_(3)^(@) = (n_(1)E_(1)^(@)F - n_(2)E_(2)^(@)F)/(nF)` `= (0.339 xx 2 - 1 xx 0.153)/(1) = 0.525 V`

Discussion

348.	Which statement is reue in regard to a spontaneous redox reaction ?A. `E_(red)^@` is always negativeB. `E_(cell)^@` is always positiveC. `E_(OX)^@` is always positiveD. `E_(cell)^@` is always positive
Answer» Correct Answer - B `DeltaG^@ =- nFE^@ cell` if `E_(cell)^@` is positive, then `Delta G^@` will be `-ve` showing htat cell reaction is spontaneous.

Discussion

349.	A standard hydrogen electrode has zero electrode potential because :A. hydrogen is easiest tio oxidiseB. this electrode potential is assumed to be zeroC. hydrogen atom has only one electronD. hydrogen is the lightest element
Answer» Correct Answer - B It is a fact.

Discussion

350.	The increase in the equivalent conductance of a salt solution on dilution is due to increase in theA. attraction between the ionsB. degree of ionization of the saltC. molecular attractionD. association of the salt
Answer» Correct Answer - B

Discussion

Explore topic-wise InterviewSolutions in .

Cell reactiomn is spontaneous whenA. `E_"red"^@` is negativeB. `DeltaG^@` is negativeC. `E_"oxide"^@` is positiveD. `DeltaG^@` is positive

Find the reduction potential of `AsO_(4)^(3-)|AsO_(2)^(c-)` in a solution when `18mL` of `0.1 N` solution of `NaI` is added to `20mL` of `0.1N Na_(3)AsO_(4)` solution at `pH=5.` The standard reduction potential of `AsO_(4)^(3-)|AsO_(2)^(c-)=-0.70V`.

If in a galvanic cell, the cell reaction is reversed as `Cu(s)|Cu^(2_(aq)||Zn^(2)aq|Zn` the cell potential will beA. 11.1VB. `-11.1V`C. `-1.1V`D. `1.1V`

Calculate the potential of the following cell `:` `Pt(s)|{:(Ce^(3+),(2M)) ,(Ce^(4+),(1M)):}||{:(Cr^(3+)(2M),) ,(Cr_(2)O_(7)^(2-)(1M),H^(o+)(1M)):}|Pt(s)` Given `:E^(c-)._(Ce^(3+)|Ce^(4+))=-1.7,E^(c-)._(Cr_(2)O_(7)^(2-)|Cr^(3+))=1.3V` `(`Take`0.059~~0.06)`

Calculate the potential of the following cell `:` `Pt|underset((2.0M)(1.0M))(Co^(2+),Co^(3+))|underset((1.0M)(4.0M)(1.0M))(Cr^(3+),Cr_(2)O_(7)^(2_-),H^(o+))|Pt` `E^(c-)._(Co^(2+)|Co^(3+))=-2V,E^(c-)._(Cr_(2)O_(7)^(2-)|Cr^(3+))=+1.0V`A. `1.024V`B. `-1.024V`C. `0.976V`D. `-0.976V`

Calculate the potential corresponding to the following cell. Given `Pt|Co^(2+)(2.0M),Co^(3+)(0.01M)||Cr^(3+)(0.5M),Cr_(2)O_(7)^(2-)(4.0M),H^(o+)(1.5M)|Pt ` `E^(c-)._(Co^(2+)|Co^(3+))=-1.82V` `E^(c-)._(Cr_(2)O_(7)^(2-)|Cr^(3+))=+1.33V`

Calculate the reduction potential of the following electrodes `:` `a.` `Pt,H_(2)(4 atm)|H_(2)SO_(4)(0.01M)` `b.` `Pt,H_(2)(1 atm)| HCl (0.2 M)` `c.` Calculate the potential of hydrogen electrode in contact with a solution whose `i pH=5" "ii. pOH=4`

The potential the cell at `25^(@)C` is Given `pK_(b)` of `NH_(4)OH=4.74`A. `0.05V`B. `-0.05V`C. `-0.28V`D. `0.28V`

Given the following cell at `25^(@)C` What will be the potential of the cell ? Given `pK_(a) ` of `CH_(3)COOH=4.74`A. `-0.42V`B. `0.42V`C. `-0.19V`D. `0.19V`

A certain quantity of electricity is passed through an aqueous solution of `AgNO_(3)` and cupric salt solution connected in series . The amount of Ag deposited is 1.08 g. then the amount of copper deposited is (At. Mass Cu = 63.5 , Ag = 108) .A. 0.6454 gB. 6.354 gC. 0.3177 gD. 3.177 g

The potential of the cell at `25^(@)C ` is Given `pK_(a)` of `CH_(3)COOH` and `pK_(b)` of `NH_(4)OH=4.74`A. `-0.04V`B. `0.04V`C. `-0.189V`D. `0.189V`

To deposit `0.6354 g` of copper by electrolysis of aqueous cupric sulpbate solution, the amount of electricity required (in coulmmbs) is.A. 9650B. 4825C. 3860D. 1930

Four alkali metals A,B,C and D are having respectively standard electrode potential as -3.05, -1.66, -0.40 and 0.80. Which one will be the most reactive.A. AB. BC. CD. D

A cell, `Ag|Ag^(+)||Cu^(2+)|Cu`, initially contains 1 M `Ag^(+)` and `1M" "Cu^(2+)` ions. Calculate the change in cell potential after the passage of 9.65 A of a current of 1 h.

Which pair of electrolytes could not be distinguished by the products of electrolysis using inert electrodes?A. 1 M `CuSO_(4)` solution, 1 M `CuCl_(2)` solutionB. 1 M KCl solution, 1 M KI solutionC. 1 M `AgNO_(3)` solution, 1 M `Cu(NO_(3))_(2)` solutionD. 1 M KCl solution, 1 M NaCl solution

Which pair of electrolytes could not be distinguished by the products of electrolysis using inert electrodes ?A. 1 M `CuSO_(4)` solution, 1 M `CuCl_(2)` solutionB. 1 M KCl solution, 1 M KCl solutionC. 1 M `AgNO_(3)` solution, 1 M NaCl solutionD. 1 M KCl solution, 1 M NaCl solution

Which pair of electrolytes could not be distinguished by the products of electrolysis using inert electrodes.A. 1M `CuSO_(4)` solution, `1M CuCl_(2)` solutionB. 1M KCl solution, 1M KI solutionC. 1 M `AgNO_(3)` solution `1M Cu(NO_(3))` solutionD. 1M `CuBr_(2)` solution `1M CuSO_(4)` solution.

The electrode potentials for `Cu^(2+)+e^(-)toCu^(+)` and `Cu^(+)e^(-)toCu_(s)` are `+0.15`V and `+0.50V` respectively the value of `E_((Cu^(2+))/(Cu))^(@)` will be?A. 0.150VB. 0.500VC. 0.325VD. 0.650V

Limiting molar conductivity of `NH_4 OH` I e., `Lambda_m (NH_4 OH)` is equal to .A. `Lambda_m (NH_4 OH) + Lambda_m(NH_4Cl)-Lambda_m(HCl)`B. `Lambda_m (NH_4 OH) + Lambda_m(Na_OH)-Lambda_m(HCl)`C. `Lambda_m (NH_4 OH) + Lambda_m(NsCl)-Lambda_m(HCl)`D. `Lambda_m (NH_4 OH) + Lambda_m(NaCl)-Lambda_m(HCl)`

`E^(@)` for `Cl_2`(g) `+2e^(-)to2Cl^-`(aq) is 1.36 V, `E^(@)` for `Cl^-` (aq) `to1//2cl_2`(g)`+e^-` is:A. 1.36VB. -1.36VC. -0.68D. 0.68V

The electrode potentials for `Cu^(2+)+e^(-)toCu^(+)` and `Cu^(+)e^(-)toCu_(s)` are `+0.15`V and `+0.50V` respectively the value of `E_((Cu^(2+))/(Cu))^(@)` will be?A. 0.650VB. 0.150VC. 0.500VD. 0.325V

Standard electrode potential of three metal `X, Y` and `Z` are `-1.2 V, +0.5 V` and `-3.0 V` respectively. The reducing power of these metals will be:A. `Xgt Ygt Z`B. `Ygt Zgt X`C. `Ygt Xgt Z`D. `Zgt Xgt Y`

The reduction potential of hydrogen half cell will be negative if :A. `p(H_(2))=1 "atm and" [H^(+)]=1.0M`B. `p(H_(2))=2 "atm and"[H^(+)]=1.0M`.C. `p(H_(2))=2"atm and"[H^(+)]=2.0M`D. `p(H_(2))=1 "atm and" [H^(+)]=2.0M`

Standard electrode potential of three metal `X, Y` and `Z` are `-1.2 V, +0.5 V` and `-3.0 V` respectively. The reducing power of these metals will be:A. `ZgtXgtY`B. `XgtYgtZ`C. `YgtZgtX`D. `YgtXgtZ`.

If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the correct relationships for the values of `DeltaG^(ɵ)` and `K_(eq)`?A. ` DeltaG^@ gt 0 , K(eq) lt 1`B. ` Delta G^@ gt 0K_(eq) , K)(eq) gt 1`C. ` Delta G^@ lt 0, K_(eq) gt 1`D. ` Zgt X gt Y`

If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the correct relationships for the values of `DeltaG^(ɵ)` and `K_(eq)`?A. `DeltaG^(@) gt 0,K_(eq) lt 1`B. `DeltaG^(@) gt 0,K_(eq) gt 1`C. `DeltaG^(@) lt 0,K_(eq) gt 1`D. `DeltaG^(@) lt 0,K_(eq) lt 1`

If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the correct relationships for the values of `DeltaG^(ɵ)` and `K_(eq)`?A. `DeltaG^(@)lt0,k_(eq)lt1`B. `DeltaG^(@)gt0,k_(eq)lt1`C. `DeltaG^(@)gt0,k_(eq)gt1`D. `DeltaG^(@)lt0,k_(eq)gt1`.

`E_(Cell)^(Theta)=1.1V` for Daniell cell. Which of the following expressions are correct description of state of equilibrium in this cell?A. `1.1=K_(c)`B. `(2.303RT)/(2F)logK_(c)=1.1`C. `logK_(c)=(2.2)/(0.059)`D. `logK_(c)=1.1`

Class 12 Chemistry MCQ Questions of Electrochemistry with Answers?

If standard reduction potentials for Pb, K, Zn and Cu are -0.126 V, -2.925 V, -0.763 V and 0.337 V, the decreasing order of reducing power is :(a) Zn > Pb > K > Cu (b) Cu > Pb > Zn > K (c) K > Zn > Pb > Cu (d) K > Pb > Cu > Zn

The standard reduction potential of Pb and Zn electrodes are `-0.126` and `-0.763 V` respectively. The cell equation will beA. `Pb^(2+)+Zn rarr Pb+Zn^(2+)`B. `Pb^(4+)+2 Zn rarr Pb+2 Zn^(2+)`C. `Zn^(2+)+Pb rarr Zn+Pb^(2+)`D. None of the above

The `E^(@)` value of Zn is -0.76 V while that of Cu is +0.34 V. Do these values help in locating the relative positions of the electrodes in the electrochemical series ?

What iis electrochemical series ? How does it help in predicting whether a particular redox reaction is feasible in a given direaction or not.

The cell constant of a conductivity cellA. change with change of electrolyteB. changes with change of concentration of electrolyteC. changes with temperture of electrolyteD. remains constant for a cell

The standard reduction potential of `Cu^(2+)//Cu` and `Cu^(2+)//Cu^(+)` are `0.339 V` and `0.153 V` respectively. The standard electrode potential of `Cu^(+)//Cu` half cell is :A. `0.525 V`B. `0.827 V`C. `0.184 V`D. `0.490 V`

Which statement is reue in regard to a spontaneous redox reaction ?A. `E_(red)^@` is always negativeB. `E_(cell)^@` is always positiveC. `E_(OX)^@` is always positiveD. `E_(cell)^@` is always positive

A standard hydrogen electrode has zero electrode potential because :A. hydrogen is easiest tio oxidiseB. this electrode potential is assumed to be zeroC. hydrogen atom has only one electronD. hydrogen is the lightest element

The increase in the equivalent conductance of a salt solution on dilution is due to increase in theA. attraction between the ionsB. degree of ionization of the saltC. molecular attractionD. association of the salt

Explore topic-wise InterviewSolutions in .

Cell reactiomn is spontaneous whenA. `E_"red"^@` is negativeB. `DeltaG^@` is negativeC. `E_"oxide"^@` is positiveD. `DeltaG^@` is positive

Find the reduction potential of `AsO_(4)^(3-)|AsO_(2)^(c-)` in a solution when `18mL` of `0.1 N` solution of `NaI` is added to `20mL` of `0.1N Na_(3)AsO_(4)` solution at `pH=5.` The standard reduction potential of `AsO_(4)^(3-)|AsO_(2)^(c-)=-0.70V`.

If in a galvanic cell, the cell reaction is reversed as `Cu(s)|Cu^(2_(aq)||Zn^(2)aq|Zn` the cell potential will beA. 11.1VB. `-11.1V`C. `-1.1V`D. `1.1V`

Calculate the potential of the following cell `:` `Pt(s)|{:(Ce^(3+),(2M)) ,(Ce^(4+),(1M)):}||{:(Cr^(3+)(2M),) ,(Cr_(2)O_(7)^(2-)(1M),H^(o+)(1M)):}|Pt(s)` Given `:E^(c-)._(Ce^(3+)|Ce^(4+))=-1.7,E^(c-)._(Cr_(2)O_(7)^(2-)|Cr^(3+))=1.3V` `(`Take`0.059~~0.06)`

Calculate the potential of the following cell `:` `Pt|underset((2.0M)(1.0M))(Co^(2+),Co^(3+))|underset((1.0M)(4.0M)(1.0M))(Cr^(3+),Cr_(2)O_(7)^(2_-),H^(o+))|Pt` `E^(c-)._(Co^(2+)|Co^(3+))=-2V,E^(c-)._(Cr_(2)O_(7)^(2-)|Cr^(3+))=+1.0V`A. `1.024V`B. `-1.024V`C. `0.976V`D. `-0.976V`

Calculate the potential corresponding to the following cell. Given `Pt|Co^(2+)(2.0M),Co^(3+)(0.01M)||Cr^(3+)(0.5M),Cr_(2)O_(7)^(2-)(4.0M),H^(o+)(1.5M)|Pt ` `E^(c-)._(Co^(2+)|Co^(3+))=-1.82V` `E^(c-)._(Cr_(2)O_(7)^(2-)|Cr^(3+))=+1.33V`

Calculate the reduction potential of the following electrodes `:` `a.` `Pt,H_(2)(4 atm)|H_(2)SO_(4)(0.01M)` `b.` `Pt,H_(2)(1 atm)| HCl (0.2 M)` `c.` Calculate the potential of hydrogen electrode in contact with a solution whose `i pH=5" "ii. pOH=4`

The potential the cell at `25^(@)C` is Given `pK_(b)` of `NH_(4)OH=4.74`A. `0.05V`B. `-0.05V`C. `-0.28V`D. `0.28V`

Given the following cell at `25^(@)C` What will be the potential of the cell ? Given `pK_(a) ` of `CH_(3)COOH=4.74`A. `-0.42V`B. `0.42V`C. `-0.19V`D. `0.19V`

A certain quantity of electricity is passed through an aqueous solution of `AgNO_(3)` and cupric salt solution connected in series . The amount of Ag deposited is 1.08 g. then the amount of copper deposited is (At. Mass Cu = 63.5 , Ag = 108) .A. 0.6454 gB. 6.354 gC. 0.3177 gD. 3.177 g

The potential of the cell at `25^(@)C ` is Given `pK_(a)` of `CH_(3)COOH` and `pK_(b)` of `NH_(4)OH=4.74`A. `-0.04V`B. `0.04V`C. `-0.189V`D. `0.189V`

To deposit `0.6354 g` of copper by electrolysis of aqueous cupric sulpbate solution, the amount of electricity required (in coulmmbs) is.A. 9650B. 4825C. 3860D. 1930

Four alkali metals A,B,C and D are having respectively standard electrode potential as -3.05, -1.66, -0.40 and 0.80. Which one will be the most reactive.A. AB. BC. CD. D

A cell, `Ag|Ag^(+)||Cu^(2+)|Cu`, initially contains 1 M `Ag^(+)` and `1M" "Cu^(2+)` ions. Calculate the change in cell potential after the passage of 9.65 A of a current of 1 h.

Which pair of electrolytes could not be distinguished by the products of electrolysis using inert electrodes?A. 1 M `CuSO_(4)` solution, 1 M `CuCl_(2)` solutionB. 1 M KCl solution, 1 M KI solutionC. 1 M `AgNO_(3)` solution, 1 M `Cu(NO_(3))_(2)` solutionD. 1 M KCl solution, 1 M NaCl solution

Which pair of electrolytes could not be distinguished by the products of electrolysis using inert electrodes ?A. 1 M `CuSO_(4)` solution, 1 M `CuCl_(2)` solutionB. 1 M KCl solution, 1 M KCl solutionC. 1 M `AgNO_(3)` solution, 1 M NaCl solutionD. 1 M KCl solution, 1 M NaCl solution

Which pair of electrolytes could not be distinguished by the products of electrolysis using inert electrodes.A. 1M `CuSO_(4)` solution, `1M CuCl_(2)` solutionB. 1M KCl solution, 1M KI solutionC. 1 M `AgNO_(3)` solution `1M Cu(NO_(3))` solutionD. 1M `CuBr_(2)` solution `1M CuSO_(4)` solution.

The electrode potentials for `Cu^(2+)+e^(-)toCu^(+)` and `Cu^(+)e^(-)toCu_(s)` are `+0.15`V and `+0.50V` respectively the value of `E_((Cu^(2+))/(Cu))^(@)` will be?A. 0.150VB. 0.500VC. 0.325VD. 0.650V

`E^(@)` for `Cl_2`(g) `+2e^(-)to2Cl^-`(aq) is 1.36 V, `E^(@)` for `Cl^-` (aq) `to1//2cl_2`(g)`+e^-` is:A. 1.36VB. -1.36VC. -0.68D. 0.68V

The electrode potentials for `Cu^(2+)+e^(-)toCu^(+)` and `Cu^(+)e^(-)toCu_(s)` are `+0.15`V and `+0.50V` respectively the value of `E_((Cu^(2+))/(Cu))^(@)` will be?A. 0.650VB. 0.150VC. 0.500VD. 0.325V

Standard electrode potential of three metal `X, Y` and `Z` are `-1.2 V, +0.5 V` and `-3.0 V` respectively. The reducing power of these metals will be:A. `Xgt Ygt Z`B. `Ygt Zgt X`C. `Ygt Xgt Z`D. `Zgt Xgt Y`

The reduction potential of hydrogen half cell will be negative if :A. `p(H_(2))=1 "atm and" [H^(+)]=1.0M`B. `p(H_(2))=2 "atm and"[H^(+)]=1.0M`.C. `p(H_(2))=2"atm and"[H^(+)]=2.0M`D. `p(H_(2))=1 "atm and" [H^(+)]=2.0M`

Standard electrode potential of three metal `X, Y` and `Z` are `-1.2 V, +0.5 V` and `-3.0 V` respectively. The reducing power of these metals will be:A. `ZgtXgtY`B. `XgtYgtZ`C. `YgtZgtX`D. `YgtXgtZ`.

If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the correct relationships for the values of `DeltaG^(ɵ)` and `K_(eq)`?A. ` DeltaG^@ gt 0 , K(eq) lt 1`B. ` Delta G^@ gt 0K_(eq) , K)(eq) gt 1`C. ` Delta G^@ lt 0, K_(eq) gt 1`D. ` Zgt X gt Y`

If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the correct relationships for the values of `DeltaG^(ɵ)` and `K_(eq)`?A. `DeltaG^(@) gt 0,K_(eq) lt 1`B. `DeltaG^(@) gt 0,K_(eq) gt 1`C. `DeltaG^(@) lt 0,K_(eq) gt 1`D. `DeltaG^(@) lt 0,K_(eq) lt 1`

If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the correct relationships for the values of `DeltaG^(ɵ)` and `K_(eq)`?A. `DeltaG^(@)lt0,k_(eq)lt1`B. `DeltaG^(@)gt0,k_(eq)lt1`C. `DeltaG^(@)gt0,k_(eq)gt1`D. `DeltaG^(@)lt0,k_(eq)gt1`.

`E_(Cell)^(Theta)=1.1V` for Daniell cell. Which of the following expressions are correct description of state of equilibrium in this cell?A. `1.1=K_(c)`B. `(2.303RT)/(2F)logK_(c)=1.1`C. `logK_(c)=(2.2)/(0.059)`D. `logK_(c)=1.1`

Class 12 Chemistry MCQ Questions of Electrochemistry with Answers?

If standard reduction potentials for Pb, K, Zn and Cu are -0.126 V, -2.925 V, -0.763 V and 0.337 V, the decreasing order of reducing power is :(a) Zn &gt; Pb &gt; K &gt; Cu (b) Cu &gt; Pb &gt; Zn &gt; K (c) K &gt; Zn &gt; Pb &gt; Cu (d) K &gt; Pb &gt; Cu &gt; Zn

The standard reduction potential of Pb and Zn electrodes are `-0.126` and `-0.763 V` respectively. The cell equation will beA. `Pb^(2+)+Zn rarr Pb+Zn^(2+)`B. `Pb^(4+)+2 Zn rarr Pb+2 Zn^(2+)`C. `Zn^(2+)+Pb rarr Zn+Pb^(2+)`D. None of the above

The `E^(@)` value of Zn is -0.76 V while that of Cu is +0.34 V. Do these values help in locating the relative positions of the electrodes in the electrochemical series ?

What iis electrochemical series ? How does it help in predicting whether a particular redox reaction is feasible in a given direaction or not.

The cell constant of a conductivity cellA. change with change of electrolyteB. changes with change of concentration of electrolyteC. changes with temperture of electrolyteD. remains constant for a cell

The standard reduction potential of `Cu^(2+)//Cu` and `Cu^(2+)//Cu^(+)` are `0.339 V` and `0.153 V` respectively. The standard electrode potential of `Cu^(+)//Cu` half cell is :A. `0.525 V`B. `0.827 V`C. `0.184 V`D. `0.490 V`

Which statement is reue in regard to a spontaneous redox reaction ?A. `E_(red)^@` is always negativeB. `E_(cell)^@` is always positiveC. `E_(OX)^@` is always positiveD. `E_(cell)^@` is always positive

A standard hydrogen electrode has zero electrode potential because :A. hydrogen is easiest tio oxidiseB. this electrode potential is assumed to be zeroC. hydrogen atom has only one electronD. hydrogen is the lightest element

The increase in the equivalent conductance of a salt solution on dilution is due to increase in theA. attraction between the ionsB. degree of ionization of the saltC. molecular attractionD. association of the salt

If standard reduction potentials for Pb, K, Zn and Cu are -0.126 V, -2.925 V, -0.763 V and 0.337 V, the decreasing order of reducing power is :(a) Zn > Pb > K > Cu (b) Cu > Pb > Zn > K (c) K > Zn > Pb > Cu (d) K > Pb > Cu > Zn