3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 49 InterviewSolution

2401.	Which of the following liberate hydrogen on reaction with dilute `H_(2)SO_(4)`A. FeB. CuC. AlD. Hg
Answer» Correct Answer - C `2Al+dil.H_(2)SO_(4)toAl_(2)SO_(4)+H_(2)uarr`.

Discussion

2402.	For a reaction `A(s)+2B^(o+) rarr A^(2+)+2B` `K_(c)` has been found to be `10^(12)`. The `E^(c-)._(cell)` isA. `0.354V`B. `0.708V`C. `0.0098V`D. `1.36V`
Answer» Correct Answer - a `E=E^(c-)-(0.059)/(2) log K` At equilibrium, `E=0` `E^(c-)=(0.059)/(2) log K =(0.059)/(2) log 10^(12)` `=(0.059)/(2)xx12=0.354`

Discussion

2403.	Zinc can be coated on iron to produce galvanised iron but the reverse is not possible. It is becauseA. zinc is lighter than ironB. zinc has lower melting point than ironC. zinc has lower negative electrode potential than ironD. zinc has higher negative electrode potential than iron
Answer» Correct Answer - D Zinc has higher negative electrode potential than iron, so iron cannot be coated on zinc.

Discussion

2404.	Galvanized iron sheets are coated withA. CopperB. NickelC. ZincD. Carbon
Answer» Correct Answer - c Factual statement

Discussion

2405.	Galvanized iron sheets and coated withA. NickelB. ChromiumC. CopperD. Zinc.
Answer» Correct Answer - D G.I. sheets are coated with zinc .

Discussion

2406.	Galvanized iron sheets are coated withA. `Cr`B. `Zn`C. `Ni`D. `Cu`
Answer» Correct Answer - B Galvanized iron is sheet iron coated with a layer of `Zn` to prevent corrosion, usually made by dipping the sheet `Al` added to prevent the formation of a brittle `Zn-Fe` alloy, Thus, galvanized iron is sometimes called Zinc-plated iron. Zinc is more easily oxidized than iron: `Zn(s) rarr Zn^(2+)(aq.)+2e^(-) E^(@) = +0.76V` So even if a scratch exposes the iron, the `Zn` is still attacked In this case, the `Zn` metal scrves as the anode and the iron is the cathode.

Discussion

2407.	Three ions sheets have been coated separately with three metals (A,B and C) whose standard electrode potentials are given below: `{:("Metal",A,B,C,"Iron"),(E_("value")^(@),-0.46,-0.66V,-0.20V,-0.44V):}` Identify in which case rusting will take place faster when coating is damaged.
Answer» Comparing oxidation potential of iron with those of metals A,B and C, iron has higher oxidation potential than C only. Hence, when coating of C is broken, rusting will become faster.

Discussion

2408.	There iron sheets have been coated separately with three metals (A,B and C ) whose standard electrode potentials are given below. `{:("Metal"," "A," "B," "C," "Iron),(,-0.46" V",-0.66" V",-0.20" V",-0.44" V"):}` Identify in which case rusting will take place faster when coating is damaged.
Answer» Rusting of iron will occur faster when coated with metal C because it is placed above iron in the activity series. Iron will take part in redox reaction and not the metal C.

Discussion

2409.	`E^(@)` of `Mg^(2+) \|\| Mg, Zn^(2+) \|\| Zn`, and `Fe^(2+)\|\|Fe` are -2.37V, -0.76V and -0.44 V respectively. Which of the following is correct?A. Mg oxidises FeB. Zn oxidises `Fe`C. Zn reduces `Mg^(2+)`D. Zn reduces `Fe^(2+)`
Answer» Correct Answer - D `E_(op(Zn))^(@) gt E_(op(Fe))^(@)` Thus `Zn to Zn^(2+) + 2e` and `Fe^(2+)+ 2e to Fe`

Discussion

2410.	For how long 2.5 ampere of current is passed to supply 5400C of charge?A. 1hrB. 2.5hrC. 6hrD. 9 hr.
Answer» Correct Answer - C `Q=1t` `t=(Q)/(I)=(54000)/(2.5"amp")=(21600)/(60xx60)` hours = 6 hours.

Discussion

2411.	How many coulombs of electricity are required for complete oxidation of 90 g of `H_(2)O`?
Answer» 90g of `H_(2)O=(90)/(18)" moles"=5"moles"` 1 mole of `H_(2)O` requires electricity=2F `therefore`5moles of `H_(2)O` will require electricity`=5xx2F=10F=10xx96500=965000C` (ii) 1000 mL of 1 M `KMnO_(4)` solution contain `KMnO_(4)=1` mol `therefore`100 mL of 0.1 M `KMnO_(4)` solution contain `KMnO_(4)=(1)/(1000)xx100xx01mol=001`mol 1 mole of `KMnO_(4)` requires electricity=5F `therefore001" mol of "KMnO_(4)` will require electricity=`001xx5F=005F=005xx96500C=4825C`

Discussion

2412.	How much charge is required for the following reaction? (i) 1mol of `Al^(3+)` to Al. (ii) 1 mol of `Cu^(2+)` to Cu. (iii) 10 mole of `MnO_(4)^(-)` to `Mn^(2+)`
Answer» (i) The electrode reaction is: `Al^(3+)+3e^(-)toAl` `therefore` Quantity of charge required for reduction of 1 mole of `Al^(3+)=3F=3xx96500C=289500C`. (ii) The electrode reaction is: `Cu^(2+)+2e^(-)toCu` `therefore`Quantity of charge required for reduction of 1 mol of `Cu^(2+)=2` faradays`=2xx96500C=193000C` ltBrgt (iii) The electrode reaction is `MnO_(4)^(-)toMn^(2+)`, i.e., `Mn^(7+)5e^(-)toMn^(2+)` `therefore`Quantity of charge required `=5F=5xx96500C=482500C`.

Discussion

2413.	A current of 2.6 ampere was passed through `CuSO_(4)` solution for 380 sec . The amount of Cu deposited is (atomic mass of Cu 63.5)A. `0.3250 g`B. `0.635` gC. `6.35` gD. `3.175 g`
Answer» Correct Answer - A `W = (E xxi xx t)/(96500) = (63.5 xx 2 .6 xx380)/(2 xx 96500) = 0.325` g

Discussion

2414.	How many coulombs are required for the oxidation of `1 mol` of `H_(2)O` to `O_(2)`?A. `9.65xx10^(4)C`B. `4.825xx10^(5)C`C. `1.93xx10^(5)C`D. `1.93xx10^(4)C`
Answer» Correct Answer - C `H_(2)Orarr1//2O_(2)+2H^(+)+2e^(-)` 1 mole 2 mole(2F) `=2xx96500C` `=1.93xx10^(5)C`

Discussion

2415.	How many coulombs are required for the oxidation of 1 mol of `H_(2)O_(2)` ?A. `9.65 xx 10^(4) C`B. `93000 C`C. `1.93 xx 10^(5) C`D. `19.3 xx 10^(2) C`
Answer» Correct Answer - C `H_(2) O_(2) rarr O_(2) + 2H^(+) + 2e^(-)` 1 mol `H_(2) O_(2) -= 2 mol e^(-)` `-= 2 xx 96500 C` `= 1.93 xx 10^(5) C`

Discussion

2416.	The charge in coulombs on 1 g ion of `N^(3-)` isA. 96500B. `2.89xx10^(5)`C. `1.45xx10^(6)`D. `6xx10^(23)`
Answer» Correct Answer - B One `N^(3-)` ion has 3 units of electric charge `therefore` 1 g ion of `N^(3-)` has `3xx6.023xx10^(23)` electrons or 1 g ion of `N^(3-)` has `3xx6.023xx10^(23)` electrons or 1 g ion `N^(3-)` has `3xx96500` coulomb charge `=289500=2.89xx10^(6)C`

Discussion

2417.	How many atoms of hydrogen are liberated at cathode , when 965 coulombs of charge is passed through water ?A. `6.02 xx 10^(21)`B. `6.02 xx 10^(23)`C. `6.02 xx 10^(-19)`D. `6.02 xx 10^(19)`
Answer» Correct Answer - A `H_(2)O to H_(2) + (1)/(2) O_(2) or 2 H^(+) + 2e^(-) to 2H` i.e., 96500 C produce 1 mole H-atoms = `6.02 xx 10^(23)` atoms . So 965 C will produce H-atoms = `6.02 xx 10^(21)`.

Discussion

2418.	Calculate the charge in coulombs required for the oxidation of: (i) 2 moles of `H_(2)O` to `O_(2)`
Answer» (i) The electrode reaction for 1 mole of `H_(2)O` is: `H_(2)O to H_(2)+(1)/(2)O_(2), i.e., O^(2-) to (1)/(2)O_(2)+2e^(-)` or `2H^(+)+2e^(-)toH_(2)` or `H_(2)Oto2H^(+)+(1)/(2)O_(2)+2e^(-)` `therefore` Quantity of electricity required for oxidation of 1 mol of `H_(2)O=2F` or Quantity of electricity required for oxidation of 2 moles of `H_(2)=4F=4xx96500C=386000C` (ii) The electrode reaction for 1 mole of FeO is: `FeOto(1)/(2)Fe_(2)O_(3), i.e., Fe^(2+) to Fe^(3+)+e^(-)` `therefore` Quantity of electricity required`=1F=96500C`.

Discussion

2419.	When a current of 0.75 A is passed through `CuSO_(4)` solution of 25 min, 0.369g of copper is deposited at the cathode. Calculate the atomic mass of copper.
Answer» Quantity of electricity passed `=(0.75A)(25xx60s)=1125C` `Cu^(2+) to 2e^(-)toCu` Thus, 1 mole of Cu (i.e., gram atomic mass) is deposited by 2F, i.e, `2xx96500C` Now, `1125` C deposit Cu=0.369 g `2xx96500C` will deposit Cu=`(0.369)/(1125)xx2xx96500g=63.3g` hence, atomic mass of copper=63.3

Discussion

2420.	Statement : At the end of electrolysis using platinum electrodes, an aqueous solution of copper sulphate turn colourless. Explanation : Copper in copper sulphate is converted to copper hydroxide during the electrolysis.A. `S` is correct but `E` is wrongB. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`.
Answer» Correct Answer - A During electrolysis `CuSO_(4)` converted into `Cu`.

Discussion

2421.	Calculate the quantity of electricity that would be required to reduce 12.3g of nitrobenzene to aniline, if the current efficiency for the process is 50%. If the potential drop across the cell is 3.0V, how much energy will be consumed?
Answer» Correct Answer - 347.40kJ

Discussion

2422.	Resistance of ` 0.2 M` solution of an electrolue is ` 50 Omega`. The specific conductance of the solution is ` 1.4 S m^^(-1)`. The resistance of `0.5` M solution of the same electrolyte is `280. Omega`. The molar conducitivity of ` 0.5 M` solution of the electrolyte is ` S m^2 "mol"^(-1)` is.A. ` 5 xx10^(-4)`B. `5 xx10^(-3)`C. `5 xx10^3`D. `5 xx10^2`
Answer» Correct Answer - A For `0.2 M` solution , ` R= 50 Omega` ` k= 14 S m^(-1) = 1.4 xx 10^(-2) S cm^(-1)` The resistivity of the solution is related to specific cnductance by ` rho = 1/k = 1/(1.4 xx 10^(-2)) Omega cm` It is known that , `R= rho 1/a` It is kinown tht , `R= rho = 1/a` `rArr l/a = R xx 1/(rho) = 50 xx 1.4 xx 10^(-2) cm` for `0.5 M` solution, `R= 280 Omega` ` k=? ` `l/a = 50 xx 1.4 xx 10^(-2) xm` The specific conductance of the solution is given by ` k= 1/(rho) = 1/R xx l/a` ` k= 1/(280) xx 50 xx 1.4 xx 10^(-2)` `2.5 xx 10^(-3) S cm^(-1)` Now , the molar conductivity of the solution is given by `Lambda_m = (K xx 1000)/M = (2.5 xx 10^(-3) xx 1000)/(0.5)` ` 5.5 cm^2 "mol"^(-1)` ` = 5 xx 10^(-4) S m^2 "mol"^(-1)` Hence, the molar conductivity of `0.5 M` solution of the electrolyte is `5xx 10^(-4) Sm^2 "mol"^(-1)`.

Discussion

2423.	The molar conducatance of `Ba^(2+)` and `Cl^(-)` are `127` and `76 ohm^(-1) cm^(-1) mol^(-1)` respectively at infinite dilution. The equivalent conductance of `BaCl_(2)` at infinte dilution will beA. 101.5B. 139.5C. 203.5D. 279.5
Answer» Correct Answer - B `lamda^(oo)BaCl_(2)=(1)/(2)lamda^(oo)Ba^(2+)+lamda^(oo)Cl^(-)` `=(127)/(2)+76=139.5ohm^(-1)cm^(-1)eq^(-1)`.

Discussion

2424.	How many atoms of calcium will be deposited from a solution of `CaCl_(2)` by a current of 5mA flowig for 60s?A. `4.68xx10^18`B. `4.68xx10^15`C. `4.68xx 10^12`D. `4.68xx10^9`
Answer» Correct Answer - A Quantity of electricity passed`=(25)/(1000)xx60=1.5` `2F=2xx96500C` deposit `Ca=`1mole `therefore1.5`C will deposit Ca=1mole `therefore1.5C` will deposit `Ca=(1)/(2xx96500)xx1.5`mole `=(1)/(2xx96500)xx1.5xx6.023xx10^(23)`atom`=4.68xx10^(18)`

Discussion

2425.	The molar conducatance of `Ba^(2+)` and `Cl^(-)` are `127` and `76 ohm^(-1) cm^(-1) mol^(-1)` respectively at infinite dilution. The equivalent conductance of `BaCl_(2)` at infinte dilution will beA. `139.5"ohm"^(-1) "cm"^(2) "eq"^(-1)`B. `203"ohm"^(-1) "cm"^(2) "eq"^(-1)`C. `279"ohm"^(-1) "cm"^(2) "eq"^(-1)`D. `101.5"ohm"^(-1) "cm"^(2) "eq"^(-1)`
Answer» Correct Answer - A `BaCl_2 to Ba^(2+) + 2Cl^(-)` `Lambda_(BaCl_2)^@ = Lambda_(Ba^(2+))^@ + 2Lambda_(Cl^-)^@`=127 + ( 2 x 76 ) `279 "ohm"^(-1) "cm"^(2) eq^(-1)` Equivalent conductivity =`279/2`=139.5 `"ohm"^(-1) "cm"^(2) eq^(-1)`

Discussion

2426.	The molar conducatance of `Ba^(2+)` and `Cl^(-)` are `127` and `76 ohm^(-1) cm^(-1) mol^(-1)` respectively at infinite dilution. The equivalent conductance of `BaCl_(2)` at infinte dilution will beA. 139.52B. 203C. 279D. 101.5
Answer» Correct Answer - A (a)The equivalent conductance of `BaCl_(2)` at infinite dilution. `lambda_(oo) " of " BaCl_(2)=(1)/(2) lambda_(oo) " of " Ba^(2+) + lambda_(oo) " of " Cl^(-)` `=(127)/(2)+76=139.5`

Discussion

2427.	The molar conducatance of `Ba^(2+)` and `Cl^(-)` are `127` and `76 ohm^(-1) cm^(-1) mol^(-1)` respectively at infinite dilution. The equivalent conductance of `BaCl_(2)` at infinte dilution will beA. `330 Omega^(-1) cm^(2)`B. `203 Omega^(-1) cm^(2)`C. `139.5 Omega^(-1) cm^(2)`D. `51 Omega^(-1) cm^(2)`
Answer» Correct Answer - C `Lambda_(m)^(oo)` for `BaCl_(2)=Lambda_(m)^(oo) Ba^(2+)+2Lambda_(m)^(oo)Cl^(-)` `:. Lambda_(eq)^(oo)` for `BaCl_(2)=1/2 Lambda_(m)^(oo)Ba^(2+)+Lambda_(m)^(oo)Cl^(-)` `=1/2 xx127+76=139.5 Omega^(-1) cm^(2)`

Discussion

2428.	The molar conducatance of `Ba^(2+)` and `Cl^(-)` are `127` and `76 ohm^(-1) cm^(-1) mol^(-1)` respectively at infinite dilution. The equivalent conductance of `BaCl_(2)` at infinte dilution will beA. `139.52`B. `203`C. `279`D. `101.5`
Answer» Correct Answer - A According to Kohlrausch law of independent migration of ions, limiting equivalent conductance of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte, Thus `Lambda_(eq.)^(@)(BaCl_(2))=lamda_(eq.)^(@)(Ba^(2+))+lamda_(eq.)^(@)(Cl^(-))` Since `Ba^(+)` is divalent while `Cl^(-)` is univalent, `lamda_(eq.)^(@)(Ba^(2+)) = 1//2` `lamda_(m)^(@)(Ba^(2+)` while `lamda_(eq.)^(@)(Cl^(-)) = lamda_(m)^(@)(Cl^(-))`. Thus `lamda_(eq.)^(@)(BaCl^(2)) = ((127)/(2)+76) ohm^(-1) cm^(-1) eq^(-1)` `=139.5 ohm^(-1) cm^(-1) eq^(-1)`

Discussion

2429.	The equilibrium constant (K) for the reaction `Cu(s)+2Ag^(+) (aq) rarr Cu^(2+) (aq)+2Ag(s)`, will be [Given, `E_(cell)^(@)=0.46 V`]A. `K_(C) = AL (15.6)`B. `K_(C) = AL(2.5)`C. `K_(C) = AL (1.5)`D. `K_(C) = AL (12.2)`
Answer» Correct Answer - A `E^(@) = (0.0592)/(n) "log" K_(C)` `therefore log K_(C) = (E^(@) xx n)/(0.0592) = (0.46 xx 2 )/(0.0592) = 15.6` `therefore K_(C) = AL (15.6)`

Discussion

2430.	The emf of the cell, `Pt\|H_(2)(1atm)\|H^(+) (0.1M, 30mL)\|\|Ag^(+) (0.8M)Ag` is `0.9V`. Calculate the emf when `40mL` of `0.05M NaOH` is added.
Answer» Correct Answer - `0.95V` `E = E^(@) - 0.0591 log.([H^(+)])/([Ag^(+)])` `0.9 - E^(@) - 0.0591 log.(0.1)/(0.8)` `E^(@) = 0.84662` On adding `40mL` of `0.05M NaOH` `(0.05M NaOH 40 mL)` ltbr. `[H^(+)] = (3-2)/(70) = (1)/(70)` now `E = E^(@) - 0.0591 log.([H^(+)])/([Ag^(+)])` `E = 0.84662 - 0.0591 log.(1)/(70 xx 0.8)` `E = 0.95V`

Discussion

2431.	The standard reduction potential for `Fe^(2+)//Fe and Sn^(2+)//Sn` electrode are -0.44 and -0.14 volt respectively. For the givencell reaction `Fe^(2+)+SntoFe+Sn^(2+)`, the standard EMF is:-A. `+0.30V`B. `-0.58V`C. `+0.58V`D. `-0.30V`
Answer» Correct Answer - D For the cell reaction, Fe acts as cathode as Sn as anode. Hence, `E_(cell)^(o)=E_(cathode)^(o)-E_("anode")^(o)=-0.44-(-0.14)=-0.30V` the negative EMF suggests that the reaction goes spontaneously in reverse direction.

Discussion

2432.	The standard reduction potential for `Fe^(2+) //Fe` and `Sn^(2+) //Sn` electrodes are `-0.44` and `-0.14 ` volt respectively. For the given cell reaction `Fe^(2+) + Sn rarr Fe + Sn^(2+) `, the standard ` EMF` is.A. ` +.030 V`B. `-0.58 V`C. ` +0. 58 V`D. ` -0. 30 V`
Answer» Correct Answer - D For the cell reaction. Fe acts as cathode and Sn as anode, Hence, `E_(cell)^2 -E_("cathode")^2 -E_("anode")^@ =- 0.44-(-0.14) =- 0.30 V` The negative EMF suggests that the reaction goes spontaneously in reversed direction.

Discussion

2433.	The standard reduction potential for `Fe^(2+) //Fe` and `Sn^(2+) //Sn` electrodes are `-0.44` and `-0.14 ` volt respectively. For the given cell reaction `Fe^(2+) + Sn rarr Fe + Sn^(2+) `, the standard ` EMF` is.A. `+.030v`B. `-0.58v`C. `+0.58v`D. `-0.300`
Answer» Correct Answer - D The two half cell reactions are `SnrarrSn^(2+)+2e^ (-)` `Fe^(2+)+2e^(-)rarrFe` `therefore` cell representation is `Sn\|Sn^(2+)\|\|Fe^(2+)\|Fe` `E_("cell")^(@)=E_(Fe^(2+)//Fe)^(@)-E_(Sn^(2+)//Sn)^(@)` `=-0.44V-(-0.14V)=0.30V`

Discussion

2434.	The standard reduction potential for `Fe^(2+) //Fe` and `Sn^(2+) //Sn` electrodes are `-0.44` and `-0.14 ` volt respectively. For the given cell reaction `Fe^(2+) + Sn rarr Fe + Sn^(2+) `, the standard ` EMF` is.A. ` -0 . 30 V`B. ` -0 . 58 V`C. ` +0. 58 V`D. ` + 0. 3 0 V`
Answer» Correct Answer - A For the cell reaction.Fe acts as cathode and Sn as anode, Hence, `E_("cell")^2 -E_("cathode")^2 -E_("anode")^@ =- 0.44-(-0.14) =- 0.30 V` The negative EMF suggests that the reaction goes spontaneously in reversed direction.

Discussion

2435.	Given, standard electrode potentials `Fe^(2+) +2e^(-) rarr Fe, E^(@)=-0.440 V` `Fe^(3+)+3e^(-) rarr Fe, E^(@)=- 0.036 V` The standarde potential `(E^(@))` for `Fe^(2+)+e^(-) rarr Fe^(2+)`, isA. `+0.772 V`B. `-0.772 V`C. `+0.417 V`D. `-0.417 V`
Answer» Correct Answer - A `Fe^(2+)+2e^(-) rarr Fe, DeltaG^(@)=-nFE^(@)` …(i) `DeltaG^(@)=-2xxFxx(-0.440V)=0.880 F` `Fe^(3+)+3e^(-) rarr Fe` …(ii) `DeltaG^(@)=-3xxFxx(-0.036)` `=0.108 F` On subtracting Eq. (ii) from Eq. (i) we get `Fe^(3+) +e^(-) rarr Fe^(2+)` `DeltaG^(@)=0.108 F-0.880 F=-0.772 F` `E^(@)=(-DeltaG^(@))/(nF)=((-0.772 F))/(1xxF)= +0.772 V`

Discussion

2436.	Given standard `E^(c-):` `Fe^(3+)+3e^(-)rarrFe," "E^(c-)=-0.036` Fe^(2+)+2e^(-)rarr Fe," "E^(c-)=-0.440V` The `E^(c-)` of `Fe^(3+)+e^(-) rarr Fe^(2+) ` isA. `-0.476V`B. `-0.404V`C. `0.404V`D. `0.772V`
Answer» Correct Answer - d `Fe^(3+)+3e^(-) rarr Fe(E_(1))" "(i)` `Fe^(2+)+2e^(-) rarr Fe(E_(2))" "(ii)` Net equation `Fe^(3+)+e^(-) rarr Fe^(2+)(E_(3))" "(iii)` `[` Obtained by `Eqs. (i)-(ii)]` `E_(3)=(n_(1)E_(1)-N_(2)e_(2))/(n_(3))=(3(-0.036)-2(-0.0440))/(1)` `=0.772V`

Discussion

2437.	The standard reduction for the following reactions are : `Fe^(3+) + 3e^(-) rarr Fe` with `E^(@) = - 0.036 V` `Fe^(2+) + 2e^(-) rarr Fe` with `E^(@) = - 0.44 V` What would be the standard electrode potential for the reaction `Fe^(3+) + e^(-) rarr Fe^(2+)` ?A. 0.772 VB. 0.077 VC. `-0.404 V`D. `0.772 V`
Answer» Correct Answer - A `{:(Fe^(3+)+3e^(-) rarr Fe,E_(1)^(@)=-0.036 V" "(n_(1)=3)),(Fe" "rarr Fe^(2+)+2e^(-),E_(2)^(@)=+0.44 V" "(n_(2)=2)),(bar(Fe^(3+)+e^(-) rarr Fe^(2+)" "),E_(3)^(@)=?" "(n_(3)=1)):}` `n_(3)E_(3)^(@)=n_(1)E_(1)^(@)+n_(2)E_(2)^(@)` `1xxE_(3)^(@)=3(-0.036)+2(+0.44)` `E_(3)^(@)=0.772 V`

Discussion

2438.	A hydrogen electrode X was placed in a buffer solution of sodium acetate and acetic acid in the ratio `a:n` and another hydrogen electode Y was placed in a buffer solution of sodium acetate and acetic acid in the ratio `b:a`. If reduction potential values for two cells are found to be `E_(1)` and `E_(2)` respectively w.r.t standard hydrogen electrode, the `pK_(a)` value of the acid can be given as:A. `(E_(1) + E_(2))/(0.118)`B. `(E_(2) - E_(1))/(0.118)`C. `(-E_(1) + E_(2))/(0.118)`D. `(E_(1) - E_(2))/(0.118)`
Answer» Correct Answer - A

Discussion

2439.	Given standard electode potenitals ` Fe^(2+) + 2e^(-) rarr Fe, E^@ =- 0. 44 V ` ………………(1) ` Fe^(3+) + 2e^(-) rarr Fe, E^@ =- 0. 036 V` ………….(2) The standard electrode pptential `E^@ ` for `Fe^(2+) + e rarr Fe^(2+)` is.A. ` -0. 476 V`B. ` -0. 404 V`C. `0. 404 V`D. ` +0. 772 V`
Answer» Correct Answer - D `Fe^(2+) + 2e rarr Fe, DeltaG_1^@ =- nE^2 F=- 2 F (-0.44)` `= 0.88 F` `Fe^(2+) + 2e rarr Fe, DeltaG_2^@ =- nE^2 F=- 2 F (-0.036)` ` =0.108 F` On subtracting (1) from (2) `Fe^(3+) +d^- rarr Fe^(2+)` or `Delta G =- 0. 772 F` or ` Delta G =- nE^@ F=- 0.772F` Or ` E^@ = (0.772 )/1 -.0772 V`.

Discussion

2440.	What is the effect of temperature on the rate constant ?
Answer» Most of the chemical reactions are accelerated by increased of temperature. `to` For a chemical with rise of temperature by `10^(@)C` the rate constant is nearly doubled `k=A.e^(-Ea//RT)`

Discussion

2441.	In electrolysis of `NaCl` when `Pt` electrode is taken `H_(2)` is liberated at cathode while `Hg` cathode it forms sodium amalgam becauseA. Hg ismore inert than PtB. More voltage is required to reduce `H^(+)` at Hg than at Pt.C. Na is dissolved in Hg while it does not dissolve in PtD. Conc. Of `H^(+)` ions is larger when Pt electrode is taken.
Answer» Correct Answer - B More voltage is required to reduc e `H^(+)` at Hg than at Pt.

Discussion

2442.	In electrolysis of `NaCl` when `Pt` electrode is taken `H_(2)` is liberated at cathode while `Hg` cathode it forms sodium amalgam becauseA. Hg is more inert than PtB. more voltage is required to reduce `H^(+)` at Hg than at PtC. Na is dissovled in Hg while it does not dissolved in PtD. concentration of `H^(+)` ion is larger when Pt electrode is taken
Answer» Correct Answer - B (b) Sodium Chloride in water dissociates as `NaCl iff Na^(+) +Cl^(-)` `H_(2)O iff H^(+) +OH^(-)` When electric current is passed through this solution using platinum eletrodes , `Na^(+) and H^(+)` move towards cathode whereas `Cl^(-) and OH^(-)` ions move towards anode. At cathode `H^(+)+e^(-) to H` `H+H to H_(2)` At anode `Cl^(-) to Cl+e^(-)` `Cl+Cl to Cl_(2)` If mercury is used as cathode, `H^(+)` ions are not discharged at mercury cathode because mercury has high hydrogen over voltage. `Na^(+)` ions are discharged at cathode in preference of `H^(+)` ions yielding sodium, which dissolves in mercury to form sodium amalgum.

Discussion

2443.	In electrolysis of `NaCl` when `Pt` electrode is taken `H_(2)` is liberated at cathode while `Hg` cathode it forms sodium amalgam becauseA. `Hg` is more insert than `Pt`B. More valtage is required to reduce `H^(+)` at `Hg` than at `Pt`C. `Na` is dissolved in `Hg` while it does not dissolve in `Pt-`D. Concentration of `H^(+)` ions is larger when `Pt` electrode is taken
Answer» Correct Answer - B `NaCl(aq.) rarr Na^(+)(aq.)+Ci^(-)` At cathode `Na^(+)(aq.)+e^(-) rarr Na(s)` or `2H_(2)O(l) rarr O_(2)(g)+4H^(+)(aq.)+4e^(-)` When `Pt` electrode is used, `H_(2)O` is reduced as its reduction potential is higher than that of `Na^(+)` ions. However, due to over-voltage for the liberation of `H_(2)(g)` at `Hg` electrode, the `H^(+)` ions arc not discharged at `Hg` cathode. Thus, `Na^(+)` ions are discharged at cathode (in perference of `H^(+)` ions) yielding `Na` metal, which dissolves in `Hg` to form sodium amalgam.

Discussion

2444.	Calculate emf of the following cell Cd/Cd2+ (.10 M)//H+ (.20 M)/H2 (0.5 atm)/Pt [Given E° for Cd2+ /Cd = -0.403 v]
Answer» E_cell= E°_cell – 0.0591/n Log [Cd²⁺]/ [H⁺]² E°_cell= 0 – (–.403 V) =0.403 V = 0.0403 – 0.0591/2 Log (0.10) X 0.5/(0.2)² = 0.400 V

2444.

Calculate emf of the following cell Cd/Cd2+ (.10 M)//H+ (.20 M)/H2 (0.5 atm)/Pt [Given E° for Cd2+ /Cd = -0.403 v]

Answer»

E_cell= E°_cell – 0.0591/n Log [Cd²⁺]/ [H⁺]²

E°_cell= 0 – (–.403 V) =0.403 V

= 0.0403 – 0.0591/2 Log (0.10) X 0.5/(0.2)² = 0.400 V

Discussion

2445.	The `EMF` of a galvanic cell `Pt\|H_(2)(1 atm)\|HCl(1M)\|Cl_(2)(g)\|Pt` is `1.29V`. Calculate the partial pressure of` Cl_(2)(g)`. `E^(c-)._(Cl_(2)\|Cl^(c-))=1.36V`.
Answer» Cell reaction `:` `(1)/(2)H_(2)(g)rarr 2H^(o+)+e^(-)" "E^(c-)._(o x i d)=0V` `2Cl_(2)(g)+e^(-) rarr Cl^(c-)(aq)" "E^(c-)._(red)=1.36V` `ulbar((1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)rarrH^(o+)(aq)+Cl^(c-)(aq)" "E^(c-)._(cell)=1.36V)` `E_(cell)=E^(c-)._(cell)-(0.059)/(1) log.([H^(o+)][Cl^(c-)])/([P_(H_(2))]^(1//2)[P_(Cl_(2))]^(1//2))` `1.29=1.36-(0.059)/(1)log ((1xx1)/(1xx[p_(Cl_(2))]^(1//2)))` Solve, `p_(Cl_(2))=4.24xx10^(-3)atm`

Discussion

2446.	In a lead storag battery, Pb (anode) and `PbO_(2)` (cathode) are used. Concentrated `H_(2)SO_(4)` is used as electrolyte. The battery holds 3.5 litre acid with it . In the discharge process, the density of acid fell from 1.294 to 1.139 g/mL. The sulphuric acid of density `1.294" g mL"^(-1)` is 39 % by mass and that of density 1.139 g/mL is 20% by mass. The amount of charge which the battery must have been used is :A. 9.88 FB. 8.98 FC. 8.89 FD. 7.88 F
Answer» Correct Answer - A

Discussion

2447.	The passage of current through the solution of a dcertain electrolyte results in the liberation of `H_(2)` at the cathode and chlorine at the anode. The solution in the container could be:A. `NaCl(aq.)`B. `CuCl_(2)(aq.)`C. `Kcl(aq.)`D. `MgCl_(2)(aq.)`
Answer» Correct Answer - A::C::D

Discussion

2448.	How can we practice Electrochemistry Class 12 Mock Test?
Answer» We have created the Electrochemistry Class 12 Online Mock Test for the understudies of class 12 for his or her board exams. during this mock test, you'll get a number of questions from the Electrochemistry section and others. If you would like to induce more marks in Electrochemistry Class 12 Mock Test then practice these online mock tests for Electrochemistry Chemistry designed as per the newest syllabus. Free Online Mock Test for class 12 Electrochemistry for important topics of all chapters. Students can attempt the mock tests for class 12 Electrochemistry as again and again. it helps you to clear all concepts. Practice our test regularly and check where you stand. Practice online mock tests will assist you to see your understanding and identify areas of improvement. you must practice these Electrochemistry Class 12 Online Mock Test having objective-based questions and compare your answers with the solutions provided by us. Click here to start practice: - Class 12 Electrochemistry Mock Test

2448.

How can we practice Electrochemistry Class 12 Mock Test?

Answer»

We have created the Electrochemistry Class 12 Online Mock Test for the understudies of class 12 for his or her board exams. during this mock test, you'll get a number of questions from the Electrochemistry section and others.

If you would like to induce more marks in Electrochemistry Class 12 Mock Test then practice these online mock tests for Electrochemistry Chemistry designed as per the newest syllabus. Free Online Mock Test for class 12 Electrochemistry for important topics of all chapters. Students can attempt the mock tests for class 12 Electrochemistry as again and again. it helps you to clear all concepts.

Practice our test regularly and check where you stand. Practice online mock tests will assist you to see your understanding and identify areas of improvement. you must practice these Electrochemistry Class 12 Online Mock Test having objective-based questions and compare your answers with the solutions provided by us.

Click here to start practice: - Class 12 Electrochemistry Mock Test

Discussion

2449.	`EMF` of the cell `Zn ZnSO_(4)(a =0.2)\|\|ZnSO_(4)(a_(2))\|Zn` is `-0.0088V` at `25^(@)C`. Calculate the value of `a_(2)`.
Answer» Correct Answer - `a_(2) = 0.1006M` `E_(cell) = (0.0591)/(2)log.([0.2])/([a_(2)])` `(-0.0088 xx 2)/(0.0591) =- log. ([0.2])/([a_(2)])` `10^(+0.3) = (0.2)/(a_(2))` `a_(2) = 0.1006`

Discussion

2450.	Calculate the `EMF` of a Daniel cell when the concentartion of `ZnSO_(4)` and `CuSO_(4)` are `0.001M` and `0.1M` respectively. The standard potential of the cell is `1.1V`.
Answer» Correct Answer - `E = 1.158V` `E_(cell) = (E_(cell)^(@) - (0.0591)/(2) log .([Zn^(2+)])/([Cu^(2+)]))` `= 1.1 -(0.0591)/(2)log.([10^(-3)])/([10^(-1)])` `= 1.1 +0.0591` ltbr. `=1.1591V`

Discussion

Explore topic-wise InterviewSolutions in .

Which of the following liberate hydrogen on reaction with dilute `H_(2)SO_(4)`A. FeB. CuC. AlD. Hg

For a reaction `A(s)+2B^(o+) rarr A^(2+)+2B` `K_(c)` has been found to be `10^(12)`. The `E^(c-)._(cell)` isA. `0.354V`B. `0.708V`C. `0.0098V`D. `1.36V`

Zinc can be coated on iron to produce galvanised iron but the reverse is not possible. It is becauseA. zinc is lighter than ironB. zinc has lower melting point than ironC. zinc has lower negative electrode potential than ironD. zinc has higher negative electrode potential than iron

Galvanized iron sheets are coated withA. CopperB. NickelC. ZincD. Carbon

Galvanized iron sheets and coated withA. NickelB. ChromiumC. CopperD. Zinc.

Galvanized iron sheets are coated withA. `Cr`B. `Zn`C. `Ni`D. `Cu`

Three ions sheets have been coated separately with three metals (A,B and C) whose standard electrode potentials are given below: `{:("Metal",A,B,C,"Iron"),(E_("value")^(@),-0.46,-0.66V,-0.20V,-0.44V):}` Identify in which case rusting will take place faster when coating is damaged.

There iron sheets have been coated separately with three metals (A,B and C ) whose standard electrode potentials are given below. `{:("Metal"," "A," "B," "C," "Iron),(,-0.46" V",-0.66" V",-0.20" V",-0.44" V"):}` Identify in which case rusting will take place faster when coating is damaged.

`E^(@)` of `Mg^(2+) || Mg, Zn^(2+) || Zn`, and `Fe^(2+)||Fe` are -2.37V, -0.76V and -0.44 V respectively. Which of the following is correct?A. Mg oxidises FeB. Zn oxidises `Fe`C. Zn reduces `Mg^(2+)`D. Zn reduces `Fe^(2+)`

For how long 2.5 ampere of current is passed to supply 5400C of charge?A. 1hrB. 2.5hrC. 6hrD. 9 hr.

How many coulombs of electricity are required for complete oxidation of 90 g of `H_(2)O`?

How much charge is required for the following reaction? (i) 1mol of `Al^(3+)` to Al. (ii) 1 mol of `Cu^(2+)` to Cu. (iii) 10 mole of `MnO_(4)^(-)` to `Mn^(2+)`

A current of 2.6 ampere was passed through `CuSO_(4)` solution for 380 sec . The amount of Cu deposited is (atomic mass of Cu 63.5)A. `0.3250 g`B. `0.635` gC. `6.35` gD. `3.175 g`

How many coulombs are required for the oxidation of `1 mol` of `H_(2)O` to `O_(2)`?A. `9.65xx10^(4)C`B. `4.825xx10^(5)C`C. `1.93xx10^(5)C`D. `1.93xx10^(4)C`

How many coulombs are required for the oxidation of 1 mol of `H_(2)O_(2)` ?A. `9.65 xx 10^(4) C`B. `93000 C`C. `1.93 xx 10^(5) C`D. `19.3 xx 10^(2) C`

The charge in coulombs on 1 g ion of `N^(3-)` isA. 96500B. `2.89xx10^(5)`C. `1.45xx10^(6)`D. `6xx10^(23)`

How many atoms of hydrogen are liberated at cathode , when 965 coulombs of charge is passed through water ?A. `6.02 xx 10^(21)`B. `6.02 xx 10^(23)`C. `6.02 xx 10^(-19)`D. `6.02 xx 10^(19)`

Calculate the charge in coulombs required for the oxidation of: (i) 2 moles of `H_(2)O` to `O_(2)`

When a current of 0.75 A is passed through `CuSO_(4)` solution of 25 min, 0.369g of copper is deposited at the cathode. Calculate the atomic mass of copper.

Calculate the quantity of electricity that would be required to reduce 12.3g of nitrobenzene to aniline, if the current efficiency for the process is 50%. If the potential drop across the cell is 3.0V, how much energy will be consumed?

The molar conducatance of `Ba^(2+)` and `Cl^(-)` are `127` and `76 ohm^(-1) cm^(-1) mol^(-1)` respectively at infinite dilution. The equivalent conductance of `BaCl_(2)` at infinte dilution will beA. 101.5B. 139.5C. 203.5D. 279.5

How many atoms of calcium will be deposited from a solution of `CaCl_(2)` by a current of 5mA flowig for 60s?A. `4.68xx10^18`B. `4.68xx10^15`C. `4.68xx 10^12`D. `4.68xx10^9`

The molar conducatance of `Ba^(2+)` and `Cl^(-)` are `127` and `76 ohm^(-1) cm^(-1) mol^(-1)` respectively at infinite dilution. The equivalent conductance of `BaCl_(2)` at infinte dilution will beA. 139.52B. 203C. 279D. 101.5

The molar conducatance of `Ba^(2+)` and `Cl^(-)` are `127` and `76 ohm^(-1) cm^(-1) mol^(-1)` respectively at infinite dilution. The equivalent conductance of `BaCl_(2)` at infinte dilution will beA. `139.52`B. `203`C. `279`D. `101.5`

The equilibrium constant (K) for the reaction `Cu(s)+2Ag^(+) (aq) rarr Cu^(2+) (aq)+2Ag(s)`, will be [Given, `E_(cell)^(@)=0.46 V`]A. `K_(C) = AL (15.6)`B. `K_(C) = AL(2.5)`C. `K_(C) = AL (1.5)`D. `K_(C) = AL (12.2)`

The emf of the cell, `Pt|H_(2)(1atm)|H^(+) (0.1M, 30mL)||Ag^(+) (0.8M)Ag` is `0.9V`. Calculate the emf when `40mL` of `0.05M NaOH` is added.

The standard reduction potential for `Fe^(2+)//Fe and Sn^(2+)//Sn` electrode are -0.44 and -0.14 volt respectively. For the givencell reaction `Fe^(2+)+SntoFe+Sn^(2+)`, the standard EMF is:-A. `+0.30V`B. `-0.58V`C. `+0.58V`D. `-0.30V`

The standard reduction potential for `Fe^(2+) //Fe` and `Sn^(2+) //Sn` electrodes are `-0.44` and `-0.14 ` volt respectively. For the given cell reaction `Fe^(2+) + Sn rarr Fe + Sn^(2+) `, the standard ` EMF` is.A. ` +.030 V`B. `-0.58 V`C. ` +0. 58 V`D. ` -0. 30 V`

The standard reduction potential for `Fe^(2+) //Fe` and `Sn^(2+) //Sn` electrodes are `-0.44` and `-0.14 ` volt respectively. For the given cell reaction `Fe^(2+) + Sn rarr Fe + Sn^(2+) `, the standard ` EMF` is.A. `+.030v`B. `-0.58v`C. `+0.58v`D. `-0.300`

The standard reduction potential for `Fe^(2+) //Fe` and `Sn^(2+) //Sn` electrodes are `-0.44` and `-0.14 ` volt respectively. For the given cell reaction `Fe^(2+) + Sn rarr Fe + Sn^(2+) `, the standard ` EMF` is.A. ` -0 . 30 V`B. ` -0 . 58 V`C. ` +0. 58 V`D. ` + 0. 3 0 V`

Given, standard electrode potentials `Fe^(2+) +2e^(-) rarr Fe, E^(@)=-0.440 V` `Fe^(3+)+3e^(-) rarr Fe, E^(@)=- 0.036 V` The standarde potential `(E^(@))` for `Fe^(2+)+e^(-) rarr Fe^(2+)`, isA. `+0.772 V`B. `-0.772 V`C. `+0.417 V`D. `-0.417 V`

Given standard `E^(c-):` `Fe^(3+)+3e^(-)rarrFe," "E^(c-)=-0.036` Fe^(2+)+2e^(-)rarr Fe," "E^(c-)=-0.440V` The `E^(c-)` of `Fe^(3+)+e^(-) rarr Fe^(2+) ` isA. `-0.476V`B. `-0.404V`C. `0.404V`D. `0.772V`

Given standard electode potenitals ` Fe^(2+) + 2e^(-) rarr Fe, E^@ =- 0. 44 V ` ………………(1) ` Fe^(3+) + 2e^(-) rarr Fe, E^@ =- 0. 036 V` ………….(2) The standard electrode pptential `E^@ ` for `Fe^(2+) + e rarr Fe^(2+)` is.A. ` -0. 476 V`B. ` -0. 404 V`C. `0. 404 V`D. ` +0. 772 V`

What is the effect of temperature on the rate constant ?

Calculate emf of the following cell Cd/Cd2+ (.10 M)//H+ (.20 M)/H2 (0.5 atm)/Pt [Given E° for Cd2+ /Cd = -0.403 v]

The `EMF` of a galvanic cell `Pt|H_(2)(1 atm)|HCl(1M)|Cl_(2)(g)|Pt` is `1.29V`. Calculate the partial pressure of` Cl_(2)(g)`. `E^(c-)._(Cl_(2)|Cl^(c-))=1.36V`.

The passage of current through the solution of a dcertain electrolyte results in the liberation of `H_(2)` at the cathode and chlorine at the anode. The solution in the container could be:A. `NaCl(aq.)`B. `CuCl_(2)(aq.)`C. `Kcl(aq.)`D. `MgCl_(2)(aq.)`

How can we practice Electrochemistry Class 12 Mock Test?

`EMF` of the cell `Zn ZnSO_(4)(a =0.2)||ZnSO_(4)(a_(2))|Zn` is `-0.0088V` at `25^(@)C`. Calculate the value of `a_(2)`.

Calculate the `EMF` of a Daniel cell when the concentartion of `ZnSO_(4)` and `CuSO_(4)` are `0.001M` and `0.1M` respectively. The standard potential of the cell is `1.1V`.