3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 6 InterviewSolution

251.	Calculate the standard potential for the reaction, `Hg_(2)Cl_(2)+Cl_(2) rarr 2Hg^(2+)+4Cl^(-)` Given : `{:(Hg_(2)Cl_(2)+2e^(-) rarr 2Hg+2Cl^(-),,E^(@)=0.270" volt"),(Hg_(2)^(2+) rarr 2Hg^(2+)+2e^(-),,E^(@)=-0.92" volt"),(2Hg rarr Hg_(2)^(2+)+2e^(-),,E^(@)=-0.79"volt"),(Cl_(2)+2e^(-) rarr 2Cl^(-),,E^(@)=1.36" volt"):}`
Answer» Correct Answer - `-0.08` volt First determine `E^(@)` for `Hg_(2)Cl_(2) rarr 2Hg^(2+) +2Cl^(-) +2e^(-)` electrode. It comes to -1.44 volt. This electrode is now coupled with `Cl_(2)+2e^(-) rarr 2Cl^(-)` electrode.

Discussion

252.	State Kohlrausch’s law and write mathematical expression of molar conductivity of the given solution at infinite dilution.
Answer» Statement of Kohlrausch’s law : This states that at infinite dilution of the solution, each ion of an electrolyte migrates independently of its co-ions and contributes independently to the total molar conductivity of the electrolyte, irrespective of the nature of other ions present in the solution. This law of independent migration of ions is represented as : ∧₀ = \(\lambda^0_+\) + \(\lambda^0_-\). Where ∧₀ is the molar conductivity of the electrolyte at infinite dilution or zero concentration while \(\lambda^0_+\) and \(\lambda^0_-\) are the molar ionic conductivities of cation and anion respectively at infinite dilution.

252.

State Kohlrausch’s law and write mathematical expression of molar conductivity of the given solution at infinite dilution.

Answer»

Statement of Kohlrausch’s law : This states that at infinite dilution of the solution, each ion of an electrolyte migrates independently of its co-ions and contributes independently to the total molar conductivity of the electrolyte, irrespective of the nature of other ions present in the solution.

This law of independent migration of ions is represented as :

∧₀ = \(\lambda^0_+\) + \(\lambda^0_-\).

Where ∧₀ is the molar conductivity of the electrolyte at infinite dilution or zero concentration while \(\lambda^0_+\) and \(\lambda^0_-\) are the molar ionic conductivities of cation and anion respectively at infinite dilution.

Discussion

253.	The cost of 2Rs/kWh of operating an electric motor for 10hours takes 10amp at 110V is:A. 79200 RsB. 22000RsC. 220RsD. 22Rs
Answer» Correct Answer - D

Discussion

254.	The specific conductivity of `0.1 N KCl` solution is `0.0129 Omega^(-1) cm^(-1)`. The resistane of the solution in the cell is `100 Omega`. The cell constant of the cell will beA. 1.1B. 1.29C. 0.56D. `2.80`
Answer» Correct Answer - B Specific conductivity `(k)=1/Rxx` cell constant Cell constant `=kxxR=0.0129xx100=1.29`

Discussion

255.	What is the potential of a half-cell consisting of zinc electrode in 0.01m `ZnSO_(4)` solution at `258^(@)C(E^(@)=0.763V)`A. 0.8221 VB. 8.221 VC. 0.5282 VD. 9.232 V
Answer» Correct Answer - A

Discussion

256.	Which of the following has been universally accepted as a reference electrode at all temperature and has been assigned a value of zero voltA. Graphite electrodeB. Copper electrodeC. Platinum electrodeD. Standard hydrogen electrode
Answer» Correct Answer - D

Discussion

257.	Match the followingColumn-I Complex (with coordination number of Co3+ as equal to six)Column-II Maximum molar conductivity (S cm2 mol–1)(A) CoCl3.6NH3(p) 97(B) CoCl3.5NH3(q) 0C) CoCl3.4NH3(r) 404(D) CoCl3.3NH3(s) 229
Answer» A→(r), B→(s), C→(p), D→(q) [Co(NH₃)₆] Cl₃ will give maximum number of ions(4) because of which conductivity is maximum i.e. 404. In [Co(NH₃) 3Cl₃] no ions are given Hence molar conductivity is zero. [Co(NH₃) 5Cl]Cl₂ & [Co(NH₃) 4Cl₂] Cl forms 3 & 2 ions.

257.

Match the followingColumn-I Complex (with coordination number of Co3+ as equal to six)Column-II Maximum molar conductivity (S cm2 mol–1)(A) CoCl3.6NH3(p) 97(B) CoCl3.5NH3(q) 0C) CoCl3.4NH3(r) 404(D) CoCl3.3NH3(s) 229

Answer»

A→(r), B→(s), C→(p), D→(q)

[Co(NH₃)₆] Cl₃ will give maximum number of ions(4) because of which conductivity is maximum i.e. 404.

In [Co(NH₃) 3Cl₃] no ions are given

Hence molar conductivity is zero.

[Co(NH₃) 5Cl]Cl₂ & [Co(NH₃) 4Cl₂] Cl forms 3 & 2 ions.

Discussion

258.	Show that in the case of first order reaction, the time required for `99.9%` completion of the reaction is 10 times that required for `50%` completion `(log 2 = 0.3010)`
Answer» `t_(1//2)=(0.693)/(k)` For `99.9%to a=100` `a-x =100 -99.9 =0.1` `k=(2.303)/(k)log ""(100)/(0.1)` But ` k =(0.693)/(t_(1//2))=(2.303)/(0.93)xxt_(1//2)log 1000=3.33xx t_(1//2)log 1000=3.33xxt_(1//2) xx3 =9.99xxt_(1//2)` `t_(99.9%)`is 10 times `t_(1//2)`

Discussion

259.	Zinc is coated over ion iron to prevent rusting of iron because :A. it is cheaper than ironB. `E_((Zn^(2+)//Zn))^(@) = E_((Fe^(2+)//Fe))^(@)`C. `E_((Zn^(2+)//Zn))^(@) lt E_((Fe^(2+)//Fe))^(@)`D. `E_((Zn^(2+)//Zn))^(@) gt E_((Fe^(2+)//Fe))^(@)`
Answer» Correct Answer - C

Discussion

260.	The resistance of a solution `A` is `50 ohm` and that of solution `B` is `100ohm,` both solutions are taken in the same conductivity cell. If equal volumes of solution`A` and `B` are mixed, what is the resistance of the mixture using the same cell ? `(` Assume there is no change or increase in the `prop` of `A` and `B` on mixing `)`.
Answer» Let `K_(1)` and `K_(2)` be the specific conductance of the solution A and B respectively and the cell constant of the cell be `X`. For solution A: Sp. Conductance = Conductance `xx` Cell constant `k_(1) = (1)/(50) xx X` ....(i) For solution B: Sp. conductance, `k_(2) = (1)/(100) xx X` ...(ii) When equal volumes of A and B are mixed, both the solutions get doubly diluted, hence their individual contribution towards the sp. conductance of the mixture will be `(k_(1))/(2)` and `(k_(2))/(2)` respectively and the sp. conductance of the mixture will be `(1)/(2) (k_(1)+k_(2))` `:.` For the mixture : `(1)/(2) (k_(1)+k_(2)) = (1)/(R) xx X` ...(iii) `(R` is the resistance of mixture) From equation (i),(ii) and (ii), `R = 66.67 ohm`.

Discussion

261.	The standard reduction potential of `Li^(+)//Li,Ba^(2+)//Ba,Na^(+)//Na` and `Mg^(2+)//Mg`. Are -3.05,-2.73,-2.71 and -2.37 volts respectively which one of the following is strongest oxidising agent?A. `Na^(+)`B. `Li^(+)`C. `Ba^(2+)`D. `Mg^(2+)`
Answer» Correct Answer - D A strong reducing agent is the one which has lowest red potential. The strongest oxidising agent is one which has maximum tendency to gain electron, i.e., whose `E_("cell")^(@)` is maximum.

Discussion

262.	The standard reduction potentials at `25^(@)C` of `Li^(+)//Li, Ba^(2+)//Ba, Na^(+)//Na` and `Mg^(2+)//Mg` are `-3.05, -2.73, -2.71` and `-2.37` volt respectively. Which one of the following is the strongest oxidising agent ?A. `Na^(+)`B. `Li^(+)`C. `Ba^(2+)`D. `Mg^(2+)`
Answer» Correct Answer - D Higher the reduction potential , stronger is oxidizing agent .

Discussion

263.	The emf of the cell `Zn\|Zn^(2+)(1M)\|\|Cu^(2+)\|Cu(1M)` is 1.1V. If the standard reduction potential of `Zn^(2+)\|Zn` is -0.78V, what is the oxidation potential of `Cu\|Cu^(2+)`?A. `+ 1.86` VB. `0.32` VC. `-0.32` VD. `1.86` V
Answer» Correct Answer - C `E_("cell") = [E_(OP_(Zn))^(@) + E_(RP _(Cu))^(@)] + (0.0592)/(2) log _(10) ([Cu^(2+)])/([Zn^(2+)])` `therefore 1.1 = 0.78 + E_(RP_(Cu))^(@) + (0.0592)/(2) log _(10) (1) (because log 1 = 0)` `therefore E_(RP)^(@) (Cu^(2+) \| Cu) = 1.1 - 0.78 = 0.32` `therefore E_(OP)^(@) (Cu\|Cu^(2+)) = -0.32`

Discussion

264.	The standard reduction potential of `Cu^(+)//Cu` couple is 0.34 V at `25^(@)C` calculate the reduction potential at `pH=14` for this couple. (Given `K_(sp),Cu(OH)_(2)=1.0xx10^(-19)`)A. `-0.22` VB. `+0.22` VC. `-0.34` VD. `+0.34` V
Answer» Correct Answer - A `Cu^(2+) + 2e^(-) to Cu`, `E = E^(@) - (0.0592)/(2) log_(10) (1)/([Cu^(2+)])` `= 0.34 - (0.059)/(2) log_(10) ((1)/(10^(-19)))= - 0.22V`.

Discussion

265.	The reaction `1//2H2(g)+AgCl(s) rarr H^(o+)(aq)+Cl^(c-)(aq)+Ag(s) ` occurs in the galvanic cell.A. `Pt\|H_(2)(g)\|KCl(aq.)\|AgCl(s)\|Ag(s)`B. `Pt\|H_(2)(g)\|HCl(aq.)\|AgCl(s)\|Ag(s)`C. `Pt\|H_(2)(g)\|HCl(aq.)\|AgNO_(30)(aq.)\|Ag(s)`D. `Ag\|AgCl(s)\|KCl(aq.)\|AgNO_(30)(aq.)\|Ag`
Answer» Correct Answer - D The standard hydrogen electrode consists of a plati-num electrode coated with platinum black (i.e. finely divided platinum). The electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it. The concentration of both the reduced and oxidized forms of hydrogen is main-tained at unity. This imlpies that the pressure of hudrogen fas is one bar and the concentration of hydrogen ions in the solution is one molar. Gas electrodes ultilize the phenomenon of adsorption of gas molecular by inert metals. The platinum electrode is pre-pareed by electrochemically depositing platicnum on platinum metal, which gives a large surface area for the adorption of gases.

Discussion

266.	The resist ance of 0./01N solution of an electrolyte AB at 328K is 100ohm. The specific conductance of solution is cell constant =`1cm^(-1)`A. 100ohmB. `10^(-2)ohm^(-1)`C. `10^(-2)ohm^(-1)cm^(-1)`D. `10^(2)ohm-cm`
Answer» Correct Answer - C Specific conductance =Cell constan t `xx(1)/("resistance")` `=1cm^(-1)xx(1)/(100"atm")` `=10^(-2)ohm^(-1)cm^(-1)`

Discussion

267.	Standard reducation potential at `25^(@)C` of `Li^(+)//Li`, `Ba^(2+)//Ba, Na^(+)//Na` and `Mg^(+)//Mg` are `-3.05, -2.90, -2.71` and `-2.37` volt respectively. Which one of the following is the strongest oxidising agent ?A. `Mg^(2+)`B. `Ba^(2+)`C. `Na^(+)`D. `Li^(+)`
Answer» Correct Answer - A (a) A cation having maximum (positive) value of standard reduction potential is the strongest oxidising agent. Hence, `Mg^(2+)` is the strongest oxidising agent.

Discussion

268.	An electrochemical cell is shown below `Pt, H_(2)(1 "atm")\|HCl(0.1 M)\|CH_(3)COOH(0.1 M)\|H_(2)(1 "atm")`, The emf of the cell will not be zero, becauseA. The pH of 0.1 M HCl and 0.1 M acetic acid is not the sameB. Acids used in the two compartments are differentC. E.M.F of a cell depends on the molarities of acids usedD. The temperature is constant
Answer» Correct Answer - A `E_("cell") ne 0` because `[H^(+)]` is not same in 0.1 M HCl and 0.1 `M CH_(3)COOH`.

Discussion

269.	An electrochemical cell is shown below `Pt, H_(2)(1 "atm")\|HCl(0.1 M)\|CH_(3)COOH(0.1 M)\|H_(2)(1 "atm")`, The emf of the cell will not be zero, becauseA. `pH` of `0.1 M HCl` and `0.1 M CH_(3)COOH` are not sameB. acids used in two compartments are differentC. the temperature is constantD. emf depends on molarities of acids used
Answer» Correct Answer - A It is a kind of concentration cell. The concentration of `H^(+)` ions in two half cell is different because `HCl` is a strong electrolyte while `CH_(3)COOH` is a weak electrolyte.

Discussion

270.	Calculate `Delta_(r)G^(c-)` of the reaction `:` `Ag^(o+)(aq)+Cl^(c-)(aq) rarr AgCl(s)` Given `:Delta_(f)G^(c-)._(AgCl)=-109kJ mol ^(-1)` `Delta_(f)G^(c-)_((Cl^(c-)))=-129k J mol ^(-1)` `Delta_(f)G^(c-)._((Ag^(o+)))=-77 kJ mol ^(-1)`A. `-97 kJ mol^(-1)`B. `-57kJ mol^(-1)`C. `57kJmol^(-1)`D. `97kJ mol^(-1)`
Answer» Correct Answer - b `Ag^(o+)(aq)+Cl^(c-)(aq)rarr AgCl(s)` `Delta_(r)G^(c-)=Delta_(f)G^(c-)._((AgCl))-Delta_(f)G^(c-)._((Ag^(o+)))-Delta_(v)G^(c-)._((Cl^(c-)))` `=-109-77-(-129)=-57kJ mol^(-1)`

Discussion

271.	An electrochemical cell is shown below `Pt, H_(2)(1 "atm")\|HCl(0.1 M)\|CH_(3)COOH(0.1 M)\|H_(2)(1 "atm")`, The emf of the cell will not be zero, becauseA. EMF depends on molarities of acids usedB. pH of 0.1 M HCl and 0.1 M `CH_(3)COOH` is sameC. the temperature is constantD. acids used in two compartments are different
Answer» Correct Answer - B (b)The EMF of the cell will not be zero because concentration of `H^(+)` ions in two electrolytic solution is different. Mean HCl is strong acid where, acetic acid is weak acid and gives different pH.

Discussion

272.	An electrochemical cell is shown below `Pt, H_(2)(1 "atm")\|HCl(0.1 M)\|CH_(3)COOH(0.1 M)\|H_(2)(1 "atm")`, The emf of the cell will not be zero, becauseA. The pH of 0.1M HCl and 0.1M acetic acid is not the sameB. Acids used in the two compartments are differentC. E.M.F. of a cell depends on the molarities of acids usedD. The temperature is contain.
Answer» Correct Answer - A `[H^(+)]` in two half cells are not the same since HCl and `CH_(3)COOH` dissociate t o different extent in water.

Discussion

273.	For the following cell reaction, `Ag\|Ag^(+)\|AgCl\|Cl^(Theta)\|Cl_(2),Pt` `DeltaG_(f)^(o)(AgCl)=-109kJ//mol` `DeltaG_(f)^(o)(Cl^(Theta))=-129kJ//mol` `DeltaG_(f)^(o)(Ag^(+))=78kJ//mol` `E^(o)` of the ell is:-A. `-0.60V`B. `0.60V`C. `6.0V`D. None of these
Answer» Correct Answer - A

Discussion

274.	The specific conductance of a 0.01 M solution of KCl is 0.0014 `ohm^(-1) cm^(-1) " at " 25^(@)C`. Its equivalent conductance is:A. 14B. 140C. `1.4`D. `0.14`
Answer» Correct Answer - B

Discussion

275.	An electrochemical cell is set up as follows `Pg(H_(2),1" atm")//0.1M" "HCl` `\|\|0.1M" acetic acid "//(H_(2),1" atm")Pt` E.M.F. of this cell will not be zero becauseA. The pH of 0.1 M HCl and 0.1 M acetic acid is not the sameB. Acids used in two compartments are differentC. E.M.F of a cell depends on the molarities of acids usedD. The temperature is constant
Answer» Correct Answer - A The pH of 0.1 M HCl and 0.1 M acetic acid is not the same, because HCl is a strong acid to its pH is less and `CH_(3)COOH` is a weak acid, so its pH is more.

Discussion

276.	The `EMF` of the following cellis `1.05V` at `25^(@)C:` `Pt,H_(2)(g)(1.0 atm)\|NaOH(0.1m),NaCl(0.1M)\|AgCl(s),Ag(s)` `a.` Write the cell reaction, `b.` Calculate `pK_(w)` of water.
Answer» Correct Answer - `pH=8.63` Half cell reactions are `:` `[Zn(s) rarr Zn^(2+)(aq)+2e^(-)]` at anode `[H^(o+)+e^(-)rarr (1)/(2)H_(2)]` at cathode For zinc electrode `:` `E_(Znn^(2+))=E^(c-)._(Zn\|Zn^(2+))-(0.0591)/(n)log[(Zn^(2+))/(Zn(s))]` `=0.76-(0.0591)/(2)log .(0.1)/(1)` `=0.76-(0.0591)/(2)xx(-1)` `=+0.76+0.03=0.79V` Similarly, for hydrogen electrode `E_(H^(o+)\|H_(2))=0-(0.0591)/(2)log .(1)/([H^(o+)]^(2))` or `E_(H^(o+)\|H_(2))=0-(0.0591)/(2)xx(-log[H^(o+)])` `:. E_(H^(o+)\|H_(2))=-0.0591pH` Now, `,E_(cell)=E_(Zn\|Zn^(2+))-E_(H^(o+)\|H_(2))` `:.0.28=+0.79-0.0591pH` or `0.0591pH=+0.79-0.28` or ` 0.0591pH=0.51` `:.pH=(0.51)/(0.0591)=8.63`

Discussion

277.	Given `l//a=0.5 cm^(-1), R= 50 "ohm", N= 1.0.` The equivalent conductance of the electrolytic cell is .A. 10 `ohm^(-1)cm^(2)gmeq^(-1)`B. 20 `ohm^(-1)cm^(2)gmeq^(-1)`C. 300 ohm S `cm^(2)eq^(-1)`D. 135.9S `cm^(2)eq^(-1)`
Answer» Correct Answer - A `1//a=0.5cm^(-1),R=50ohm` `p=(Ra)/(I)=(50)/(0.5)=100` `wedge=kxx(1000)/(N)=(1)/(p)xx(1000)/(N)=(1)/(100)0xx(1000)/(1)` `=10ohm^(-1)cm^(2)gm" "eq^(-1)`.

Discussion

278.	The reaction taking place in the cell `Pg\|H_(2)(g)\|HCl(1.0M)\|AgCl\|Ag` is 1 atmA. `AgCl+(1//2)H_(2)toAg+H^(+)+Cl^(-)`B. `Ag+H^(+)+Cl^(-)toAgCl+(1//2)H_(2)`C. `2Ag^(+)+H_(2)to2Ag+2H^(+)`D. `2Ag+2H^(+)to2Ag^(+)+H_(2)`.
Answer» Correct Answer - A LHS electrode (oxidation half reaction) `(1)/(2)H_(2)toH^(+)+e^(-)` RHS electroe (reduction half reaction) `AgCl+e^(-)toAg+Cl^(-)` Adding the two half ractions, we get `AgCl+e^(-)toAg+H^(+)+Cl^(-)`

Discussion

279.	Equivalent conductivity at infinite dilution for sodium potassium oxalate, `(COO^(-))_(2)Na^(+)K^(+)`, will be (given, molar conductivities of oxalate, `K^(+) and Na^(+)` ions at infinite diluton are `148.2,50.1,73.5" S "cm^(2)mol^(-1)` respectively).A. `271.8" S "cm^(2)eq^(-1)`B. `67.96" S "cm^(2)eq^(-1)`C. `543.6" S "cm^(2)eq^(-1)`D. `135.9" S "cm^(2)eq^(-1)`.
Answer» Correct Answer - D `wedge_(m)^(@)` (Sod. Pot. Oxalate)`=lamda_(m)^(@)(Na^(+))+lamda_(m)^(@)(K^(+))+lamda^(2)`(oxalate) `=(50.1+73.5+148.2)" S "cm^(2)mol^(-1)` `=271.8" S "cm^(2)mol^(-1)` `wedge_(eq)^(@)=(wedge_(m)^(@))/("Total chrge on cations or anions (n factor)")` `=(271.8)/(2)=135.9" S "cm^(2)eq^(-1)`.

Discussion

280.	Equivalent conductivity at infinite dilution for sodium potassium oxalate `(COO^(-))_(2)Na^(+)K^(+)` will be [given, molar conductivities of oxalate, `K^+ and Na^(+)` ions at infinite dilution are 148.2, 50.1, 73.5 S `cm^(2)mol^(-1)`, respectively]A. `271.8cm^(2)eq^(-1)`B. `67.95S" "cm^(2)eq^(-1)`C. `543.6S" "cm^(2)eq^(-1)`D. `135.9S" "cm^(2)eq^(-1)`
Answer» Correct Answer - D `lamda_(M)^(oo)=lamda_(M)^(oo)("Oxalate")+lamda_(M)^(oo)(Na^(+))+lamda_(M)^(oo)(k^(+))` `lamda_(M)^(oo)=(148.2+50.1+73.5)S" "cm^(2)mol^(-1)` `lamda_(M)^(oo)=271.8S" "cm^(2)mol^(-1)` `thereforelamda_(Eq)^(oo)=(271.8)/(2)=135.9S" "cm^(2)eq^(-1)(lamda_(eq)^(oo)=(lamda_(M)^(oo))/("n. factor"))`

Discussion

281.	The molar ionic conductivities of `NH_(4)^(+) annd OH^(-)` at infinite dilution are 72 and 198 `ohm^(-1)cm^(2)`, respectively, the molar conductivity of a centinormal `NH_(4)OH` solution aththe same temperature is found to be 9 `ohm^(-1)cm^(2)`. The percentage dissociation of `NH_(4)OH` at this concentration will beA. `33%`B. `7.14 %`C. `12.5%`D. `4.54 %`
Answer» Correct Answer - C `Lambda^(@) (NH_(4)OH) = lambda_(NH_(4)^(+))^(@) + lambda_(OH^(-))^(@)` `= 72 + 198 = 270 ohm^(-1) cm(2)` `alpha = Lambda//Lambda^(@) = 9//270 = 1//30 = 0.0333` `% alpha = 3.33 %`

Discussion

282.	Molar conductivities at infinite dilution of Mg2+ and Br- are 105.8 Ω cm2 mol-1 and 78.2 Ω-1 cm2 mol-1 respectively. Calculate molar conductivity at zero concentration of MgBr2.
Answer» Given : \(\lambda^0_{Mg^{2+}}\)= 105.8 Ω^-1 cm² mol^-1 \(\lambda^0_{Br^-}\)= 78.2 Ω^-1 cm² mol^-1 \(∧_{0(MgBr_2)}\)= ? By Kohlrausch’s law, \(∧_{0(MgBr_2)}\) = \(\lambda^0_{Mg^{2+}}\) + \(2\lambda^0_{Br^-}\) = 105.8 + 2 × 78.2 = 105.8 + 156.4 = 262.2 Ω^-1 cm² mol^-1 ∴ Molar conductivity of MgBr₂ at zero concentration = \(∧_{0(MgBr_2)}\) = 262.2 Ω^-1 cm² mol^-1

282.

Molar conductivities at infinite dilution of Mg2+ and Br- are 105.8 Ω cm2 mol-1 and 78.2 Ω-1 cm2 mol-1 respectively. Calculate molar conductivity at zero concentration of MgBr2.

Answer»

Given :

\(\lambda^0_{Mg^{2+}}\)= 105.8 Ω^-1 cm² mol^-1

\(\lambda^0_{Br^-}\)= 78.2 Ω^-1 cm² mol^-1

\(∧_{0(MgBr_2)}\)= ?

By Kohlrausch’s law,

\(∧_{0(MgBr_2)}\) = \(\lambda^0_{Mg^{2+}}\) + \(2\lambda^0_{Br^-}\)

= 105.8 + 2 × 78.2

= 105.8 + 156.4

= 262.2 Ω^-1 cm² mol^-1

∴ Molar conductivity of MgBr₂ at zero concentration = \(∧_{0(MgBr_2)}\)

= 262.2 Ω^-1 cm² mol^-1

Discussion

283.	Molar conductivities at infinite dilution (at 298 K) of `NH_(4)CI`, NaOH and NaCI are 129.8, 217.4 and 108.9 `Omega^(-1)cm^(-1)mol^(-1)` respcetively. If the molar conductivity of a centimolar solution of `NH_(4)OH` is 9.33 `Omega^(-1) cm^(-1)`, what is percentage dissociation of `NA_(4)OH` at this concentation ? Also calculte the dissociation constant for `NH_(4)OH`.
Answer» Correct Answer - `3.90% ; 1.57xx10^(-5) mol L^(-1)` Step I. Calcualtion of `Lambda_(m)^(oo)` of `NH_(4)OH` `Lambda_(m(NH_(4)OH))^(oo)=Lambda_(m(NH_(4)Cl))^(oo)+Lambda_(m(NaOH))^(oo)-Lambda_(m(NaCl))^(oo)` `=(129.8+217.4-108.9)" ohm"^(-1)cm^(2)mol^(-1)=238.3" ohm"^(-1)cm^(2)mol^(-1)`. Step II. Calculation of percent dissociation of `NH_(4)OH` Degree of dissociation `(alpha)=(Lambda_(m)^(C))/(Lambda_(m)^(oo))=((9.33" ohm"^(-1) cm^(2)mol^(-1)))/((238.3" ohm"^(-1)cm^(2)mol^(-1)))=0.039 or 3.9%` Step III. Calculation of dissociation constant `NH_(4)OH hArrNH_(4)^(+)(aq)+OH^(-)(aq)` `{:("Initial molar conc".:," "C,""0,""0),("Eqm.molar conc".:,C(1-alpha),Calpha,Calpha):}` `k_(C)=([NH_(4)^(+)(aq)][OH^(-)(aq)])/([NH_(4)OH^(-)])=(CalphaxxCalpha)/(C(1-alpha))=(Calpha^(2))/(1-alpha)` `C=M//100=0.01" mol "L^(-1), alpha=3.9xx10^(-2)` `:. " " k_(C)=((0.01" mol L "^(-1))xx(3.9xx10^(-2))^(2))/((1-3.9xx10^(-2)))=1.57xx10^(-5)" mol L"^(-1)`.

Discussion

284.	On the basis of information available from the reaction `(4)/(3)Al+O_(2) rarr (2)/(3)Al_(2)O_(3),DeltaG = -827kJ mol^(-1)` of `O_(2)`, the minimum emf required to carry out of the electrolysis of `Al_(2)O_(3)` is `(F=96,500 C mol^(-1))`A. `2.14 V`B. `4.28 V`C. `6.42 V`D. `8.56 V`
Answer» Correct Answer - A We are given `(4)/(3) Al+O_(2) rarr (2)/(3) Al_(2)O_(3)` For every `1` mol of `O_(2)`, we need to perform electrolysis of `(2)/(3)` mol of `Al_(2)O_(3)`: `(2)/(3) mol Al_(2)O_(3) rarr (4)/(3) Al^(3+)+2O^(2-)` At cathode `(4)/(3)Al^(3+)+4e^(-) rarr (4)/(3) Al` At anode `2O^(2-) rarr 4e^(-)+O_(2)` Since `Delta_(r)G^(@) = -nFE_("cell")^(@)` we have `E_("cell")^(@) = (Delta_(r)G^(ɵ))/(-nF)` `= (-827 xx 10^(3) J mol^(-1))/(-(4 mol) (96,500 C mol^(-1)))` `= 2.14 V`

Discussion

285.	On the basis of information available from the reaction `(4)/(3)Al+O_(2) rarr (2)/(3)Al_(2)O_(3),DeltaG = -827kJ mol^(-1)` of `O_(2)`, the minimum emf required to carry out of the electrolysis of `Al_(2)O_(3)` is `(F=96,500 C mol^(-1))`A. 8.56 VB. 2.14 VC. 4.28 VD. 6.42 V
Answer» Correct Answer - B `V=(827xx10^(3))/(4xx96500)=2.14V`

Discussion

286.	Molar conductance of `1M` solution of weak acid `HA` is `20 ohm^(-1) cm^(2) mol^(-1)`. Find `%` dissocation of `HA`: `((^^_(m)^(o)(H^(+)) =350 S cm^(2) mol^(-1)),(^^_(m)^(o)(A^(-)) =50 S cm^(2) mol^(-1)))`
Answer» Correct Answer - 5 `^^_(m)^(o) (HA) = 350 +50 = 400` `alpha =(^^_(m)^(C))/(^^_(m)^(o)) xx 100 = (20)/(400) xx100 = 5%`

Discussion

287.	The initial concentratin of `N_(2)O_(5)` in the following first order reaction `N_(2)O_(5)(g) to 2NO_(2)(g)+(1)/(2O_(2))(g) was 1.24 xx10^(-2) mol L^(-1)` at 318K. The concentration of `N_(2)O_(5)` after 60 minutes was `0.20xx10^(-2) mol L^(-1).` Calculate the rate constant of the reaction at 138 K.
Answer» For a first order reaction `log ""([R]_(1))/([R]_(2))=(k(t_(2)-t_(1)))/(2.303)` `k= (2.303)/((t_(2)- t_(1)))log ""([R_(1)])/([R]_(2))` `=(2.303)/((60 min -0min ))log ""(1.24xx10^(-2)mol L ^(-1))/(0.20xx10^(-2)mol L^(-1))` `=(2.303)/(60)log 6.2 min ^(-1)` `k=0.0304min^(-1).`

Discussion

288.	Assertion: Electrical conductivity of copper inicreases with increase in temperature. Reason: The electrical conductivity of metals is due to the motion of electrons.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 assertion is false but reason is true.
Answer» Correct Answer - D Electrical conductivity of copper decreases with increase in temperature because the metallic conductivity is due to the motion of electrons. On increasing temperature the motion of electron increases which hindered in conductance of current. Hence, here assertion is false but the reason is true.

Discussion

289.	Calculate the Avogadro's number using the charge on the electron 1.60 x 10-19 C and the fact that 96500 C deposits 107.9 g silver from its solution.
Answer» 96500 coulomb deposits 107.9 g Ag or E g Ag or A g Ag, where E and A are equivalent weight and atomic weight of Ag respectively. E = A because Ag is monovalent. Thus 96500 coulomb charge means charge on N electrons where N in Av. no. Thus N x e = 96500 N = 96500/1.60 x 10^-19 = 6.03 x 10²³ electrons

289.

Calculate the Avogadro's number using the charge on the electron 1.60 x 10-19 C and the fact that 96500 C deposits 107.9 g silver from its solution.

Answer»

96500 coulomb deposits 107.9 g Ag or E g Ag or A g Ag, where E and A are equivalent weight and atomic weight of Ag respectively.

E = A because Ag is monovalent. Thus 96500 coulomb charge means charge on N electrons where N in Av. no.

Thus N x e = 96500

N = 96500/1.60 x 10^-19

= 6.03 x 10²³ electrons

Discussion

290.	Which of the following reactions canot be a base for electrochemical cell?A. `H_2+O_2to H+_2O`B. `AgNO_3+Znto Zn(NO_3)_2+Ag`C. `AgNO_3+NaCl to AgCldarr+NaNO_2`D. `KMnO_4+FeSO_4+H_2SO_4 to K_2SO_4+Fe_2(SO_4)_3+MnSO_4+H_2O`
Answer» Correct Answer - D These are not electrolytes.

Discussion

291.	Which of the following reactions canot be a base for electrochemical cell?A. `H_2+O_2 to H_2O`B. `AgNO_3 + Zn to Zn(NO_3)_2 + Ag`C. `AgNO_3 + NaCl to AgCl darr + NaNO_3`D. `KMnO_4 + FeSO_4 + H_2SO_4 to K_2SO_4 + Fe_2(SO_4)_3 + MnSO_4 + H_2O`
Answer» Correct Answer - D These are not electrolyte

Discussion

292.	In a galvanic cell, the salt bridge (i) does not participate chemically in the cell reaction (ii) stops the diffusion of ions from one electrolytes to another (ii) is necessary for the occurrence of the cell reaction (iv) ensures mixing of the two elecrolytic solutionsA. (i) and (iii) onlyB. (i) and (ii) onlyC. (iii) and (iv) onlyD. all of these.
Answer» Correct Answer - B Salt bridge keeps the solutions in two half-cells electrically neutral. It prevents transference or diffusion of the ions from one half -cell to the other

Discussion

293.	In a galvanic cell, the salt bridge (i) does not participate chemically in the cell reaction (ii) stops the diffusion of ions from one electrolytes to another (ii) is necessary for the occurrence of the cell reaction (iv) ensures mixing of the two elecrolytic solutionsA. `(i), (ii), (iii), (iv)`B. `(i), (iii), (iv)`C. `(i), (ii),(iii)`D. `(i), (iii)`
Answer» Correct Answer - D The slat bridge in necessary for the occurrence of the cell reaction because it helps in completing the circuit and also helps in removing the charges on electrolytic solutions. However, it does not participate in the cell reaction.

Discussion

294.	In a galvanic cell, the salt bridge.A. Does not participate chemically in the cell reactionB. Stops the diffusion of ions from one electrode to anotherC. Is necessary for the occurrence of the cell reactionD. Ensures mixing of the two electrolytic solutions
Answer» Correct Answer - A::B Note: We feel option (C) is incorrect because in some type of concentration cells, salt bridge is not required. Which can be confirmed from NCERT (XII-Chemistry, Part-1) in Sub section 3.2 Galvanic Cell. "The electrolytes of the two half-cells are connected internally through a salt bridge as shown in Fig. 3.1 Sometimes, both the electrodes dip in the same electrolyte solution and in such cases we do not require a salt bridge."

Discussion

295.	For the following electrochemical cell at `298K` `Pt(s)+H_(2)(g,1bar) \|H^(+) (aq,1M)\|\|M^(4+)(aq),M^(2+)(aq)\|Pt(s)` `E_(cell) = 0.092 V` when `([M^(2+)(aq)])/([M^(4+)(aq)])=10^(@)` Guven, `E_(M^(4+)//M^(2+))^(@) = 0.151V, 2.303 (RT)/(F) = 0.059` The value of x is-A. -2B. -1C. 1D. 2
Answer» Correct Answer - D At anode: `H_(2)(g) hArr 2H^(+)(aq) +2e^(-)` At cathode: `M^(4+)(aq) +2e^(-) hArr M^(2+) (aq)` Net cell reaction: `H_(2)(g) +M^(4+)(aq) hArr` `2H^(+) (aq) +M^(2+)(aq)` Now, `E_(cell) = (E_(M^(4+)//M^(2+))^(@)-E_(H^(+)//H_(2))^(@))-(0.059)/(n)log.([H^(+)]^(2)[M^(2+)])/(P_(H_(2))[M^(4+)])` or, `0.092 = (0.151 -0) -(0.059)/(2)log.(1^(2)xx[M^(2+)])/(1xx[M^(4+)])` `:. ([M^(2+)])/([M^(4+)]) = 10^(2) rArr x = 2`

Discussion

296.	For the following electrochemical cell at 298 K `Pt(s)\|H_(2),(g,1" bar")\|H^(+)(aq,1M)\|\|M^(4+)(aq)\|M^(2+)(aq)\|Pt(s)` `E_(cell)=0.092V` when `([M^(2+)(aq)])/([M^(4+)(aq)])=10^(x)` Given `E_(M^(4+)//M^(2+))^(@)=0.151V,2.303(RT)/(F)=0.059V` the value of x isA. `-2`B. `-1`C. `1`D. `2`
Answer» Correct Answer - D The reactions occurring in the cell are At anode: `H_(2)(g)to2H^(+)(aq)+2e^(-)` `underline("At cathode: "M^(4+)(aq)+2e^(-)toM^(2+)(aq))` Overall reaction: `H_(2)(g)+M^(4+)(aq)toM^(2+)(aq)+2H^(+)(aq)` `E_(cell)=E_(cell)^(@)-(0.059)/(2)"log"([M^(2+)][H^(+)]^(2))/([M^(4+)])` `E_(cell)^(@)-E_(M^(4+)//M^(2+))^(@)-E_(H^(+)//H_(2))^(@)` `=0.151-0=0.151V` `therefore0.092=0.151-(0.059)/(2)log(10^(x)xx1^(2))` or `0.092=0.151-0.0295log10^(x)` or `0.0295log10^(x)=0.151-0.092=0.059` or `log10^(x)=(0.059)/(0.0295)=2` `therefore10^(x)`=Antilog `2=10^(2)thereforex=2`

Discussion

297.	The most convenient method to protect the bottom of the ship made of iron isA. coating it with red lead oxideB. white tin platingC. connecting it with Mg blockD. connecting it with Pb block
Answer» Correct Answer - B (b) The most convenient method to protect the bottom of ship made of iron in white tin plating which prevents the build up of barnacles.

Discussion

298.	For the following elecrochemical cell at 298K `Pt(s)\|H_(2)(g,1"bar")\|H^(+)(aq,1M)\|\|M^(4+)(aq),M^(2+)(aq)\|Pt(s)` `E_(cell)=0.092V` when `([M^(2+)(aq)])/([M^(4+)(aq)])=10^(x)` Given: `E_(M^(4+)//M^(2+))^(0)=0.151V,2.303(RT)/(F)=0.059V` the value of x isA. `-2`B. `-1`C. `1`D. `2`
Answer» Correct Answer - D Anode: `H_(2)-2ehArr2H^(+)` `underline("Cathode: "M^(4+)+2ehArrM^(2+))` Net cell reaction: `H_(2)+M^(4+)hArr2H^(+)+M^(2+)` `E_(cell)=E_(cell)^(0)-(0.059)/(2)"log"([H^(+)]^(2)[M^(2+)])/([M^(4+)]xxP_(H_(2)))` `0.092=0.151-(0.059)/(2)log10^(x)` `0.092=0.151-(0.059)/(2)x` `(0.059x)/(2)=0.151-0.092` `0.059x=0.059xx2impliesx=2`.

Discussion

299.	Cell reactiomn is spontaneous whenA. `E_(red)^(@)` is negativeB. `E_(red)^(@)` is positiveC. `DeltaG^(@)` is negativeD. `DeltaG^(@)` is positive
Answer» Correct Answer - C ( c) When the value of `DeltaG^(@)` is negative, the cell reaction is spontaneous. ` DeltaG^(@)=-nFE^(@)` where, n= number of electrons take part F=Faraday constant `E^(@)`=EMF of the cell Thus, for a spontaneous reaction, the EMF of the cell must be positive.

Discussion

300.	Cell reactiomn is spontaneous whenA. `E_("red")^(@)` is negativeB. `E_("red")^(@)` is positiveC. `DeltaG^(@)` is negativeD. `DeltaG^(@)` is positive
Answer» Correct Answer - C If the standard emf of the cell is `E` and `nF` is the amount of charge passed and `Delta_(r)G^(@)` is the standard Gibbs energy of the reaction, then `Delta_(r)G^(@) = -nFE_("cell")^(@)` For a reaction to be spontaneous, `Delta_(r)G^(@)` should be negative while `E_("cell")^(@)` should be positive.

Discussion

Explore topic-wise InterviewSolutions in .

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On the basis of information available from the reaction `(4)/(3)Al+O_(2) rarr (2)/(3)Al_(2)O_(3),DeltaG = -827kJ mol^(-1)` of `O_(2)`, the minimum emf required to carry out of the electrolysis of `Al_(2)O_(3)` is `(F=96,500 C mol^(-1))`A. `2.14 V`B. `4.28 V`C. `6.42 V`D. `8.56 V`

On the basis of information available from the reaction `(4)/(3)Al+O_(2) rarr (2)/(3)Al_(2)O_(3),DeltaG = -827kJ mol^(-1)` of `O_(2)`, the minimum emf required to carry out of the electrolysis of `Al_(2)O_(3)` is `(F=96,500 C mol^(-1))`A. 8.56 VB. 2.14 VC. 4.28 VD. 6.42 V

Molar conductance of `1M` solution of weak acid `HA` is `20 ohm^(-1) cm^(2) mol^(-1)`. Find `%` dissocation of `HA`: `((^^_(m)^(o)(H^(+)) =350 S cm^(2) mol^(-1)),(^^_(m)^(o)(A^(-)) =50 S cm^(2) mol^(-1)))`

The initial concentratin of `N_(2)O_(5)` in the following first order reaction `N_(2)O_(5)(g) to 2NO_(2)(g)+(1)/(2O_(2))(g) was 1.24 xx10^(-2) mol L^(-1)` at 318K. The concentration of `N_(2)O_(5)` after 60 minutes was `0.20xx10^(-2) mol L^(-1).` Calculate the rate constant of the reaction at 138 K.

Calculate the Avogadro's number using the charge on the electron 1.60 x 10-19 C and the fact that 96500 C deposits 107.9 g silver from its solution.

Which of the following reactions canot be a base for electrochemical cell?A. `H_2+O_2to H+_2O`B. `AgNO_3+Znto Zn(NO_3)_2+Ag`C. `AgNO_3+NaCl to AgCldarr+NaNO_2`D. `KMnO_4+FeSO_4+H_2SO_4 to K_2SO_4+Fe_2(SO_4)_3+MnSO_4+H_2O`

Which of the following reactions canot be a base for electrochemical cell?A. `H_2+O_2 to H_2O`B. `AgNO_3 + Zn to Zn(NO_3)_2 + Ag`C. `AgNO_3 + NaCl to AgCl darr + NaNO_3`D. `KMnO_4 + FeSO_4 + H_2SO_4 to K_2SO_4 + Fe_2(SO_4)_3 + MnSO_4 + H_2O`

In a galvanic cell, the salt bridge.A. Does not participate chemically in the cell reactionB. Stops the diffusion of ions from one electrode to anotherC. Is necessary for the occurrence of the cell reactionD. Ensures mixing of the two electrolytic solutions

For the following electrochemical cell at `298K` `Pt(s)+H_(2)(g,1bar) |H^(+) (aq,1M)||M^(4+)(aq),M^(2+)(aq)|Pt(s)` `E_(cell) = 0.092 V` when `([M^(2+)(aq)])/([M^(4+)(aq)])=10^(@)` Guven, `E_(M^(4+)//M^(2+))^(@) = 0.151V, 2.303 (RT)/(F) = 0.059` The value of x is-A. -2B. -1C. 1D. 2

For the following electrochemical cell at 298 K `Pt(s)|H_(2),(g,1" bar")|H^(+)(aq,1M)||M^(4+)(aq)|M^(2+)(aq)|Pt(s)` `E_(cell)=0.092V` when `([M^(2+)(aq)])/([M^(4+)(aq)])=10^(x)` Given `E_(M^(4+)//M^(2+))^(@)=0.151V,2.303(RT)/(F)=0.059V` the value of x isA. `-2`B. `-1`C. `1`D. `2`

The most convenient method to protect the bottom of the ship made of iron isA. coating it with red lead oxideB. white tin platingC. connecting it with Mg blockD. connecting it with Pb block

For the following elecrochemical cell at 298K `Pt(s)|H_(2)(g,1"bar")|H^(+)(aq,1M)||M^(4+)(aq),M^(2+)(aq)|Pt(s)` `E_(cell)=0.092V` when `([M^(2+)(aq)])/([M^(4+)(aq)])=10^(x)` Given: `E_(M^(4+)//M^(2+))^(0)=0.151V,2.303(RT)/(F)=0.059V` the value of x isA. `-2`B. `-1`C. `1`D. `2`

Cell reactiomn is spontaneous whenA. `E_(red)^(@)` is negativeB. `E_(red)^(@)` is positiveC. `DeltaG^(@)` is negativeD. `DeltaG^(@)` is positive

Cell reactiomn is spontaneous whenA. `E_("red")^(@)` is negativeB. `E_("red")^(@)` is positiveC. `DeltaG^(@)` is negativeD. `DeltaG^(@)` is positive