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Electrochemistry

JEE Chemistry · 118 questions · Page 3 of 12 · Click an option or "Show Solution" to reveal answer

Q21
Given : EFe3+/FeoE_{F{e^{3 + }}/Fe}^o = -0.036V; EFe2+/FeoE_{F{e^{2 + }}/Fe}^o = -0.439 V The value of standard electrode potential for the change, Fe3+ (aq) + e- →\to Fe2+ (aq) will be
A -0.072 V
B 0.385 V
C 0.770 V
D 0.270
Correct Answer
Option C
Solution

Given

Fe3++3e−→Fe,  E∘Fe3+/FeF{e^{3 + }} + 3{e^ - } \to Fe,\,\,{E^ \circ }_{F{e^{3 + }}/Fe}
           =−0.036V   ...(i)\,\,\,\,\,\,\,\,\,\,\, = - 0.036V\,\,\,...\left( i \right)
Fe2++2e−→Fe,  E∘Fe2+/FeF{e^{2 + }} + 2{e^ - } \to Fe,\,\,{E^ \circ }_{F{e^{2 + }}/Fe}
           =−0.439V   ...(ii)\,\,\,\,\,\,\,\,\,\,\, = - 0.439V\,\,\,...\left( {ii} \right)

we have to calculate

Fe3++e−→Fe2+,  ΔG=?F{e^{3 + }} + {e^ - } \to F{e^{2 + }},\,\,\Delta G = ?

To obtain this equation subtract equ

(ii)(ii)

from

(i)(i)

we get

Fe3++e−→Fe2+   ...(iii)F{e^{3 + }} + {e^ - } \to F{e^{2 + }}\,\,\,...\left( {iii} \right)

As we know that

ΔG=−nFE\Delta G = - nFE

Thus for reaction

(iii)(iii)
ΔG=ΔG1−ΔG; \Delta G = \Delta {G_1} - \Delta G;\,
 −nFE∘=−nFE1−(−nFE2)\, - nF{E^ \circ } = - nF{E_1} - \left( { - nF{E_2}} \right)
 −nFE∘=nFE2−nFE1\, - nF{E^ \circ } = nF{E_2} - nF{E_1}
−1FE∘=2×0.439F−3×0.036 F- 1F{E^ \circ } = 2 \times 0.439F - 3 \times 0.036\,F
−1FE∘=0.770 F- 1F{E^ \circ } = 0.770\,F

∴\therefore

    E∘=−0.770V\,\,\,\,{E^ \circ } = - 0.770V
O−−>F−>Na+>Mg++>Al3+{O^{ - - }} > {F^ - } > N{a^ + } > M{g^{ + + }} > A{l^{3 + }}
Q22
In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is CH3OH(l) + 3/2O2 →\to CO2 (g) + 2H2O (l) At 298K standard Gibb’s energies of formation for CH3OH(l), H2O(l) and CO2 (g) are -166.2, -237.2 and -394.4 kJ mol−1 respectively. If standard enthalpy of combustion of methanol is -726 kJ mol−1, efficiency of the fuel cell will be
A 87%
B 90%
C 97%
D 80%
Correct Answer
Option C
Solution
CH3OH(l)+32O2(g)C{H_3}OH\left( l \right) + {3 \over 2}{O_2}\left( g \right)
                   →CO2(g)+2H2O(l)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to C{O_2}\left( g \right) + 2{H_2}O\left( l \right)
ΔGr=ΔGf(CO2,g)+2ΔGf(H2O,l)−\Delta {G_r} = \Delta {G_f}\left( {C{O_2},g} \right) + 2\Delta {G_f}\left( {{H_2}O,l} \right) -
                    ΔGf(CH3OH,l)−32ΔGf(O2,g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {G_f}\left( {C{H_3}OH,l} \right) - {3 \over 2}\Delta {G_f}\left( {{O_2},g} \right)
=−394.4+2(−237.2)−(−166.2)−0= - 394.4 + 2\left( { - 237.2} \right) - \left( { - 166.2} \right) - 0
=−394.4−474.4+166.2= - 394.4 - 474.4 + 166.2
=−702.6 kJ= - 702.6\,kJ
%  \% \,\,

efficiency

  =702.6726×100\,\, = {{702.6} \over {726}} \times 100
=97%= 97\%
Q23
The correct order of EM2+/MoE_{{M^{2 + }}/M}^o values with negative sign for the four successive elements Cr, Mn, Fe and Co is :
A Mn > Cr > Fe > Co
B Cr > Fe > Mn > Co
C Fe > Mn > Cr > Co
D Cr > Mn > Fe > Co
Correct Answer
Option A
Solution

The value of

EM2+/M∘E_{{M^{2 + }}/M}^ \circ

for given metal ions are

EMn2+/Mn∘=−1.18 V,  E_{M{n^{2 + }}/Mn}^ \circ = - 1.18\,V,\,\,
ECr2+/Cr∘=−0.9V,E_{C{r^{2 + }}/Cr}^ \circ = - 0.9V,
EFe2+/Fe∘=−0.44 VE_{F{e^{2 + }}/Fe}^ \circ = - 0.44\,V

and

ECo2+/Co∘=−0.28 V.E_{C{o^{2 + }}/Co}^ \circ = - 0.28\,V.

So when we consider these values with a negative sign (i.e., taking the magnitude of the negative potentials), we get the values as : CrCr : 0.91 V MnMn : 1.18 V FeFe : 0.44 V CoCo : 0.28 V Therefore, the correct order of EM2+/M∘E_{M^{2+}/M}^\circ values with the negative sign will be : Mn > Cr > Fe > Co This matches with option A.

Q24
The Gibbs energy for the decomposition of Al2O3 at 500oC is as follows : 23Al2O3{2 \over 3}A{l_2}{O_3} →\to 43Al+O2{4 \over 3}Al + {O_2}, ΔrG{\Delta _r}G = + 966 kJ mol–1 The potential difference needed for electrolytic reduction of Al2O3 at 500oC is at least :
A 4.5 V
B 3.0 V
C 2.5 V
D 5.0 V
Correct Answer
Option C
Solution
ΔG=−nFE\Delta G = - nFE

or

E=ΔG−nF=966×1034×36500E = {{\Delta G} \over { - nF}} = {{966 \times {{10}^3}} \over {4 \times 36500}}
=−2.5  V= - 2.5\,\,V

∴\therefore The potential difference needed for the reduction

=2.5=2.5
VV
Q25
The reduction potential of hydrogen half cell will be negative if :
A p(H2) = 1 atm and [H+] = 1.0 M
B p(H2) = 1 atm and [H+] = 2.0 M
C p(H2) = 2 atm and [H+] =1.0 M
D p(H2) = 2 atm and [H+] =2.0 M
Correct Answer
Option C
Solution
H++e−⟶ 12H2;{H^ + } + {e^ - }\overset{\,}\longrightarrow {1 \over 2}{H_2};
             \,\,\,\,\,\,\,\,\,\,\,\,\,
E=E∘−0.0591log⁡PH21/2[H+]E = {E^ \circ } - {{0.059} \over 1}\log {{P_{{H_2}}^{{{1}/{2}}}} \over {\left[ {{H^ + }} \right]}}

Now if

PH2=2  {P_{{H_2}}} = 2\,\,

atm and

[H+]=1M\left[ {{H^ + }} \right] = 1M

then

E=0−0.0591log⁡21/21=−2E = 0 - {{0.059} \over 1}\log {{{2^{1/2}}} \over 1} = - 2
Q26
The standard reduction potentials for Zn2+/ Zn, Ni2+/ Ni, and Fe2+/ Fe are –0.76, –0.23 and –0.44 V respectively. The reaction X + Y2+ →\to X2+ + Y will be spontaneous when :
A X = Ni, Y = Fe
B X = Ni, Y = Zn
C X = Fe, Y = Zn
D X = Zn, Y = Ni
Correct Answer
Option D
Solution

For a spontaneous reaction

ΔG\Delta G

must be

−ve-ve

Since

ΔG=−nFE∘\Delta G = - nF{E^ \circ }

Hence for

ΔG\Delta G

to be

−ve-ve
ΔE∘\Delta {E^ \circ }

has to be positive. Which is possible when

X=Zn,Y=NiX = Zn,Y = Ni
Zn+Ni++⟶ Zn+++NiZn + N{i^{ + + }}\overset{\,}\longrightarrow Z{n^{ + + }} + Ni
EZn/Zn+2∘+ENi2+/Ni∘E_{Zn/Z{n^{ + 2}}}^ \circ + E_{N{i^{2 + }}/Ni}^ \circ
=0.76+(−0.23)=+0.53= 0.76 + \left( { - 0.23} \right) = + 0.53

(positive).

Q27
Given ECr2+/CroE_{C{r^{2 + }}/Cr}^o = -0.74 V; EMnO4−/Mn2+oE_{MnO_4^ - /M{n^{2 + }}}^o = 1.51 V ECr2O72−/Cr3+oE_{C{r_2}O_7^{2 - }/C{r^{3 + }}}^o = 1.33 V; ECl/Cl−oE_{Cl/C{l^ - }}^o = 1.36 V Based on the data given above, strongest oxidising agent will be :
A Cr3+
B Mn2+
C MnO4−MnO_4^ -
D Cl-
Correct Answer
Option C
Solution

In electrochemistry, the strongest oxidizing agent will be the one with the highest standard electrode potential (E°), because a higher E° value means a greater tendency to gain electrons, i.e., get reduced.

An oxidizing agent gains electrons and in doing so, oxidizes another species.

From the provided data, the species with the highest standard electrode potential (E°) is MnO₄⁻, with an E° of 1.51 V.

This means that MnO₄⁻ has the greatest tendency to gain electrons and thus is the strongest oxidizing agent.

Therefore, the correct answer is : Option C :

MnO4−MnO_4^ -

.

Q28
The equivalent conductance of NaCl at concentration C and at infinite dilution are λC{\lambda _C} and λ∞{\lambda _\infty }, respectively. The correct relationship between λC{\lambda _C} and λ∞{\lambda _\infty } is given as: (where the constant B is positive)
A λC=λ∞+(B)C{\lambda _C} = {\lambda _\infty } + (B)C
B λC=λ∞−(B)C{\lambda _C} = {\lambda _\infty } - (B)C
C λC=λ∞−(B)C{\lambda _C} = {\lambda _\infty } - (B)\sqrt C
D λC=λ∞+(B)C{\lambda _C} = {\lambda _\infty } + (B)\sqrt C
Correct Answer
Option C
Solution

According to Debye Huckle onsager equation,

                         λC=λ∞−BC\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\lambda _C} = {\lambda _\infty } - B\sqrt C
Q29
To find the standard potential of M3+/M electrode,the following cell is constituted : Pt/M/M3+(0.001 mol L−1 )/Ag+(0.01 mol L−1 )/Ag The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3+ + 3e−→ \to M at 298 K will be : (Given EAg+ / Ag−E_{A{g^ + }\,/\,Ag}^ - at 298 K = 0.80 Volt)
A 0.38 Volt
B 0.32 Volt
C 1.28 Volt
D 0.66 Volt
Correct Answer
Option B
Solution

According to Nernst equation, for the cell reaction M(s) + 3Ag+(aq) →\to M3+(aq) + 3Ag(s)

Ecell=Ecello−0.059nlog⁡[M3+][Ag+]3{E_{cell}} = E_{cell}^o - {{0.059} \over n}\log {{[{M^{3 + }}]} \over {{{[A{g^ + }]}^3}}}

Substituting the values, we get

0.421=Ecello−0.0593log⁡0.001(0.01)30.421 = E_{cell}^o - {{0.059} \over 3}\log {{0.001} \over {{{(0.01)}^3}}}
Ecello=0.48VE_{cell}^o = 0.48V

We know that

Ecello=EAg+/Ago−EM3+/MoE_{cell}^o = E_{A{g^ + }/Ag}^o - E_{{M^{3 + }}/M}^o
0.48=0.8−EM3+/Mo0.48 = 0.8 - E_{{M^{3 + }}/M}^o
EM3+/Mo=0.32VE_{{M^{3 + }}/M}^o = 0.32V
Q30
Resistance of 0.2 M solution of an electrolyte is 50 Ω\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 Ω\Omega. The molar conductivity of 0.5 M solution of the electrolyte in S m2 mol-1 is :
A 5 × 103
B 5 × 102
C 5 × 10-4
D 5 × 10-3
Correct Answer
Option C
Solution

Given for

0.20.2
MM

solution

R=50ΩR = 50\Omega
κ=1.45 S m−1=1.4×10−2 S cm−1\kappa = 1.45\,S\,{m^{ - 1}} = 1.4 \times {10^{ - 2}}\,S\,c{m^{ - 1}}

Now,

R=ρℓa=1κ×ℓaR = \rho {\ell \over a} = {1 \over \kappa } \times {\ell \over a}
⇒ℓa=R×κ=50×1.4×10−2\Rightarrow {\ell \over a} = R \times \kappa = 50 \times 1.4 \times {10^{ - 2}}

For

0.50.5
MM

solution

R=280 Ω; κ=?R = 280\,\Omega ;\,\kappa = ?
ℓa=50×1.4×10−2{\ell \over a} = 50 \times 1.4 \times {10^{ - 2}}
⇒R=ρℓa=1κ×ℓa\Rightarrow R = \rho {\ell \over a} = {1 \over \kappa } \times {\ell \over a}
⇒    κ=1280×50×1.4×10−2\Rightarrow \,\,\,\,\kappa = {1 \over {280}} \times 50 \times 1.4 \times {10^{ - 2}}
=1280×70×10−2=2.5×10−3 S cm−1= {1 \over {280}} \times 70 \times {10^{ - 2}} = 2.5 \times {10^{ - 3}}\,S\,c{m^{ - 1}}

Now,

Λm=κ×1000M=2.5×10−3×10000.5{\Lambda _m} = {{\kappa \times 1000} \over M} = {{2.5 \times {{10}^{ - 3}} \times 1000} \over {0.5}}
=5 S cm2mol−1=5×10−4 S m2 mol−1= 5\,S\,c{m^2}mo{l^{ - 1}} = 5 \times {10^{ - 4}}\,S\,{m^2}\,mo{l^{ - 1}}
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