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Electrochemistry question

2009 · Shift 0 · Q21
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Electrochemistry question

2009 · Shift 0 · Q21

JEE MainChemistryElectrochemistryMCQ+4 / −1
Given : EFe3+/FeoE_{F{e^{3 + }}/Fe}^oEFe3+/Feo​= -0.036V; EFe2+/FeoE_{F{e^{2 + }}/Fe}^oEFe2+/Feo​= -0.439 V The value of standard electrode potential for the change, Fe3+Fe^{3+}Fe3+ (aq) + e-→\to→ Fe2+Fe^{2+}Fe2+ (aq) will be
  1. A
    -0.072 V
  2. B
    0.385 V
  3. C
    0.770 V
  4. D
    0.270
View written solutionFree

Correct answer: C

  1. We are given two standard reduction potentials:

EFe3+/Fe∘=−0.036 VE^\circ_{Fe^{3+}/Fe} = -0.036\text{ V}EFe3+/Fe∘​=−0.036 V EFe2+/Fe∘=−0.439 VE^\circ_{Fe^{2+}/Fe} = -0.439\text{ V}EFe2+/Fe∘​=−0.439 V

We need to find:

Fe3++e−→Fe2+Fe^{3+} + e^- \to Fe^{2+}Fe3++e−→Fe2+

with standard electrode potential EFe3+/Fe2+∘E^\circ_{Fe^{3+}/Fe^{2+}}EFe3+/Fe2+∘​.

  1. Write the two given reduction half-reactions:

Fe3++3e−→FeE∘=−0.036 VFe^{3+} + 3e^- \to Fe \qquad E^\circ = -0.036\text{ V}Fe3++3e−→FeE∘=−0.036 V Fe2++2e−→FeE∘=−0.439 VFe^{2+} + 2e^- \to Fe \qquad E^\circ = -0.439\text{ V}Fe2++2e−→FeE∘=−0.439 V

  1. Convert these into standard Gibbs free energy relations using:

ΔG∘=−nFE∘\Delta G^\circ = -nFE^\circΔG∘=−nFE∘

For Fe3++3e−→FeFe^{3+} + 3e^- \to FeFe3++3e−→Fe

ΔG1∘=−3F(−0.036)=0.108F\Delta G_1^\circ = -3F(-0.036) = 0.108FΔG1∘​=−3F(−0.036)=0.108F

For Fe2++2e−→FeFe^{2+} + 2e^- \to FeFe2++2e−→Fe

ΔG2∘=−2F(−0.439)=0.878F\Delta G_2^\circ = -2F(-0.439) = 0.878FΔG2∘​=−2F(−0.439)=0.878F

  1. The required reaction is:

Fe3++e−→Fe2+Fe^{3+} + e^- \to Fe^{2+}Fe3++e−→Fe2+

Notice that:

(Fe3++3e−→Fe)=(Fe3++e−→Fe2+)+(Fe2++2e−→Fe)(Fe^{3+} + 3e^- \to Fe) = (Fe^{3+} + e^- \to Fe^{2+}) + (Fe^{2+} + 2e^- \to Fe)(Fe3++3e−→Fe)=(Fe3++e−→Fe2+)+(Fe2++2e−→Fe)

So,

ΔG1∘=ΔGreq∘+ΔG2∘\Delta G_1^\circ = \Delta G_{req}^\circ + \Delta G_2^\circΔG1∘​=ΔGreq∘​+ΔG2∘​

Hence,

ΔGreq∘=ΔG1∘−ΔG2∘\Delta G_{req}^\circ = \Delta G_1^\circ - \Delta G_2^\circΔGreq∘​=ΔG1∘​−ΔG2∘​

ΔGreq∘=0.108F−0.878F=−0.770F\Delta G_{req}^\circ = 0.108F - 0.878F = -0.770FΔGreq∘​=0.108F−0.878F=−0.770F

  1. For the required reaction, n=1n=1n=1, so

ΔGreq∘=−FEFe3+/Fe2+∘\Delta G_{req}^\circ = -FE^\circ_{Fe^{3+}/Fe^{2+}}ΔGreq∘​=−FEFe3+/Fe2+∘​

Thus,

−FE∘=−0.770F-F E^\circ = -0.770F−FE∘=−0.770F

E∘=0.770 VE^\circ = 0.770\text{ V}E∘=0.770 V

  1. Therefore, the standard electrode potential is:

0.770 V\boxed{0.770\text{ V}}0.770 V​

So the correct option is C.

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