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Electrochemistry

JEE Chemistry · 118 questions · Page 2 of 12 · Click an option or "Show Solution" to reveal answer

Q11
Aluminium oxide may be electrolysed at 1000oC to furnish aluminium metal (Atomic mass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is Al3+ + 3e- →\to Alo To prepare 5.12 kg of aluminium metal by this method would require
A 5.49 ×\times 107 C of electricity
B 1.83 ×\times 107 C of electricity
C 5.49 ×\times 104 C of electricity
D 5.49 ×\times 101 C of electricity
Correct Answer
Option A
Solution
11

mole of

e−=1F=96500 C{e^ - } = 1F = 96500\,C
27g27g

of

AlAl

is deposited by

3×96500 C3 \times 96500\,C
51205120
gg

of

AlAl

will be deposited by

=3×96500×512027= {{3 \times 96500 \times 5120} \over {27}}
=5.49×107C= 5.49 \times {10^7}C
Q12
Conductivity (Seimen’s S) is directly proportional to area of the vessel and the concentration of the solution in it and is inversely proportional to the length of the vessel then, then constant of proportionality is expressed in :
A Sm mol-1
B Sm2 mol-1
C S-2m2 mol
D S2m2 mol-2
Correct Answer
Option B
Solution

Given

S∝area × concℓ=κm2molm×m3S \propto {{area\, \times \,conc} \over \ell } = {{\kappa {m^2}mol} \over {m \times {m^3}}}

∴\therefore

    κ=Sm2mol−1\,\,\,\,\kappa = S{m^2}mo{l^{ - 1}}
Q13
The cell, Zn | Zn2+ (1M) || Cu2+ (1M) | Cu(EcelloE_{cell}^o = 1.10V) was allowed to be completely discharged at 298 K. The relative concentration of Zn2+ to Cu2+ [[Zn2+][Cu2+]]\left[ {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right] is
A antilog (24.08)
B 37.3
C 1037.3
D 9.65 ×\times 104
Correct Answer
Option C
Solution
Ecell=0;  {E_{cell}} = 0;\,\,

when cell is completely discharged.

Ecell=E∘cell−0.0592log⁡([Zn2+][Cu2+]){E_{cell}} = {E^ \circ }_{cell} - {{0.059} \over 2}\log \left( {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right)

or

    0−1.1−0.0592log⁡([Zn2+][Cu2+])\,\,\,\,0 - 1.1 - {{0.059} \over 2}\log \left( {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right)
log⁡([Zn2+][Cu2+])=2×1.10.059=37.3\log \left( {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right) = {{2 \times 1.1} \over {0.059}} = 37.3

∴\therefore

    ([Zn2+][Cu2+])=1037.3\,\,\,\,\left( {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right) = {10^{37.3}}
Q14
For the following cell with hydrogen electrodes at two different pressure p1 and p2. What will be the emf for the given cell : Pt(H2)∣H+(aq)∣Pt(H2)     p1              1M            p2\begin{aligned}& Pt({H_2})|{H^ + }(aq)|Pt({H_2}) \\ & \,\,\,\,\,{p_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,1M\,\,\,\,\,\,\,\,\,\,\,\,{p_2}\end{aligned}
A RTFlog⁡eP1P2{{RT} \over F}{\log _e}{{{P_1}} \over {{P_2}}}
B RT2Flog⁡eP1P2{{RT} \over 2F}{\log _e}{{{P_1}} \over {{P_2}}}
C RTFlog⁡eP2P1{{RT} \over F}{\log _e}{{{P_2}} \over {{P_1}}}
D none of these
Correct Answer
Option B
Solution

Oxidation half cell : -

H2(g)⟶ 2H+(1M)+2e−   ...P1{H_2}\left( g \right)\overset{\,}\longrightarrow 2{H^ + }\left( {1M} \right) + 2{e^ - }\,\,\,...{P_1}

Reduction half cell

2H+(1M)+2e−⟶ H2(g)   ...P22{H^ + }\left( {1M} \right) + 2{e^ - }\overset{\,}\longrightarrow {H_2}\left( g \right)\,\,\,...{P_2}

The net cell reaction

H2P1(g)⟶ H2P2(g)\mathop {{H_2}}\limits_{{P_1}} \left( g \right)\overset{\,}\longrightarrow \mathop {{H_2}}\limits_{{P_2}} \left( g \right)
Ecell∘=0.00 VE_{cell}^ \circ = 0.00\,V
     \,\,\,\,\,
n=2n=2

∴\therefore

    \,\,\,\,
Ecell=Ecell∘−RTnFlog⁡cK{E_{cell}} = E_{cell}^ \circ - {{RT} \over {nF}}{\log _c}K
=0−RTnFlog⁡eP2P1= 0 - {{RT} \over {nF}}{\log _e}{{{P_2}} \over {{P_1}}}

or

    Ecell=RT2Flog⁡eP1P2\,\,\,\,{E_{cell}} = {{RT} \over {2F}}{\log _e}{{{P_1}} \over {{P_2}}}
Q15
Which of the following reaction is possible at anode?
A 2Cr3+ + 7H2O →Cr2O72−\to Cr_2O_7^{2-} + 14H+
B F2 →\to 2F-
C (1/2) O2 + 2H+ →\to H2O
D none of these
Correct Answer
Option A
Solution
2Cr3++7H2O→Cr2O72−+14H+2C{r^{3 + }} + 7{H_2}O \to C{r_2}O_7^{2 - } + 14{H^ + }
O.S.  O.S.\,\,

of

CrCr

changes from

+3+3

to

+6+6

by loss of electrons. At anode oxidation takes place.

Q16
Standard reduction electrode potentials of three metals A,B&C are respectively +0.5 V, -3.0 V & -1.2 V. The reducing, powers of these metals are
A A > B > C
B C > B > A
C A > C > B
D B > C > A
Correct Answer
Option D
Solution
ABC+0.5C−3.0V−1.2V\begin{array}{lll}A & B & C \\ { + 0.5C} & { - 3.0V} & { - 1.2V} \end{array}

NOTE : The higher the negative value of reduction potential, the more is the reducing power. Hence

B>C>A.B > C > A.
Q17
For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be 0.295 V at 25oC. The equilibrium constant of the reaction at 25oC will be
A 29.5 ×\times 10-2
B 10
C 1 ×\times 1010
D 1 ×\times 10-10
Correct Answer
Option C
Solution

The equlibrium constant is released to the standard emf of cell by the expression

log⁡ K=E∘cell×n0.059\log \,K = {E^ \circ }_{cell} \times {n \over {0.059}}
=0.295×20.059= 0.295 \times {2 \over {0.059}}
log⁡ K=59059=10\log \,K = {{590} \over {59}} = 10

or

     K=1×1010\,\,\,\,\,K = 1 \times {10^{10}}
Q18
Given the data at 25oC, Ag + I- →\to AgI + e- , Eo = 0.152 V Ag →\to Ag+ + e-, Eo = -0.800 V What is the value of log Ksp for AgI? (2.303 RT/F = 0.059 V)
A –8.12
B +8.612
C –37.83
D –16.13
Correct Answer
Option D
Solution
(i)   Ag→Ag++e−  \left( i \right)\,\,\,Ag \to A{g^ + } + {e^ - }\,\,
                     E∘=−0.800 V\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = - 0.800\,V
(ii)   Ag+I−→AgI+e−\left( {ii} \right)\,\,\,Ag + {{\rm I}^ - } \to Ag{\rm I} + {e^ - }
                     E∘=0.152 V\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = 0.152\,V

From

(i)(i)

and

(ii)(ii)

we have,

AgI→Ag++I−Ag{\rm I} \to A{g^ + } + {{\rm I}^ - }
               E∘=−0.952  V\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = - 0.952\,\,V
Ecell∘=0.059nlog⁡ KE_{cell}^ \circ = {{0.059} \over n}\log \,K

∴\therefore

−0.952=0.0591 log⁡[Ag+][I−]- 0.952 = {{0.059} \over 1}\,\log \left[ {A{g^ + }} \right]\left[ {{{\rm I}^ - }} \right]

[ as

   k=[Ag+][I−]\,\,\,k = \left[ {A{g^ + }} \right]\left[ {{{\rm I}^ - }} \right]

or

   −0.9520.059=log⁡ Ksp\,\,\, - {{0.952} \over {0.059}} = \log \,{K_{sp}}

or

   −16.13=log⁡ Ksp\,\,\, - 16.13 = \log \,{K_{sp}}
Q19
Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100Ω\Omega . The conductivity of this solution is 1.29 S m–1. Resistance of the same cell when filled with 0.2 M of the same solution is 520 Ω\Omega , The molar conductivity of 0.02 M solution of the electrolyte will be
A 124 ×\times 10–4 S m2 mol–1
B 1240 ×\times 10–4 S m2 mol–1
C 1.24 ×\times 10–4 S m2 mol–1
D 12.4 ×\times 10–4 S m2 mol–1
Correct Answer
Option D
Solution
R=100Ω,κ=1R(la),laR = 100\Omega ,\kappa = {1 \over R}\left( {{l \over a}} \right),{l \over a}

(cell constant)

                 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
                 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
=1.29×100m−1= 1.29 \times 100{m^{ - 1}}

Given,

R=520Ω,C=0.2M, μR = 520\Omega ,C = 0.2M,\,\mu

(molar conductivity) == ?

μ=κ×V        \mu = \kappa \times V\,\,\,\,\,\,\,\,

(κ\kappa can be calculated as

κ=1R(1a)\kappa = {1 \over R}\left( {{1 \over a}} \right)

now cell constant is known. Hence,

μ=1520×129×10000.2×10−6m3\mu = {1 \over {520}} \times 129 \times {{1000} \over {0.2}} \times {10^{ - 6}}{m^3}
=12.4×10−4 Sm2mol−1= 12.4 \times {10^{ - 4}}\,S{m^2}mo{l^{ - 1}}
Q20
Given ECr3+/CroE_{C{r^{3 + }}/Cr}^o = -0.72 V; EFe2+/FeoE_{Fe^{2+}/Fe}^o = -0.42V, The potential for the cell Cr | Cr3+ (0.1M) || Fe2+ (0.01 M) | Fe is
A 0.26 V
B 0.399 V
C −0.339 V
D −0.26 V
Correct Answer
Option A
Solution

From the given representation of the cell,

Ecell{E_{cell}}

can be found as follows.

Ecell=EFe2+/Fe∘−ECr3+/Cr∘−0.0596log⁡[Cr3+]2[Fe2+]3{E_{cell}} = E_{F{e^{2 + }}/Fe}^ \circ - E_{C{r^{3 + }}/Cr}^ \circ - {{0.059} \over 6}\log {{{{\left[ {C{r^{3 + }}} \right]}^2}} \over {{{\left[ {F{e^{2 + }}} \right]}^3}}}
[\left[ {} \right.

Nernst - Equ.

]\left. {} \right]
=−0.42−(−0.72)−0.0596log⁡(0.1)2(0.01)3= - 0.42 - \left( { - 0.72} \right) - {{0.059} \over 6}\log {{{{\left( {0.1} \right)}^2}} \over {{{\left( {0.01} \right)}^3}}}
=−0.42+0.72−0.0596log⁡0.1×0.10.01×0.01×0.01= - 0.42 + 0.72 - {{0.059} \over 6}\log {{0.1 \times 0.1} \over {0.01 \times 0.01 \times 0.01}}
=0.3−0.0596log⁡10−210−6= 0.3 - {{0.059} \over 6}\log {{{{10}^{ - 2}}} \over {{{10}^{ - 6}}}}
=0.3−0.0596×4= 0.3 - {{0.059} \over 6} \times 4
=0.30−0.0393=0.26 V= 0.30 - 0.0393 = 0.26\,V

Hence option (d) is correct answer.

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