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Electrochemistry question

2017 · 8 Apr · Shift 1 · Q14
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Electrochemistry question

2017 · 8 Apr · Shift 1 · Q14

JEE MainChemistryElectrochemistryMCQ+4 / −1
What is the standard reduction potential (Eo) for Fe3+Fe^{3+}Fe3+ →\to→ Fe ? Given that : Fe2+Fe^{2+}Fe2+ + 2e −→- \to−→ Fe; EFe2+/FeoE_{F{e^{2 + }}/Fe}^oEFe2+/Feo​=−-− 0.47 V Fe3+Fe^{3+}Fe3+ + e −→- \to−→ Fe2+Fe^{2+}Fe2+; EFe3+/Fe2+oE_{F{e^{3 + }}/F{e^{2 + }}}^oEFe3+/Fe2+o​ = +0.77 V
  1. A
    −-− 0.057 V
  2. B
    + 0.057 V
  3. C
    + 0.30 V
  4. D
    −-− 0.30 V
View written solutionFree

Correct answer: A

  1. We need the standard reduction potential for:

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

  1. Given half-reactions:

Fe3++e−→Fe2+,E1∘=+0.77 VFe^{3+} + e^- \rightarrow Fe^{2+}, \qquad E_1^\circ = +0.77\text{ V}Fe3++e−→Fe2+,E1∘​=+0.77 V

Fe2++2e−→Fe,E2∘=−0.47 VFe^{2+} + 2e^- \rightarrow Fe, \qquad E_2^\circ = -0.47\text{ V}Fe2++2e−→Fe,E2∘​=−0.47 V

  1. These two half-reactions add to give the required reaction:

Fe3++e−→Fe2+Fe^{3+} + e^- \rightarrow Fe^{2+}Fe3++e−→Fe2+ Fe2++2e−→FeFe^{2+} + 2e^- \rightarrow FeFe2++2e−→Fe [4pt] \Rightarrow Fe^{3+} + 3e^- \rightarrow Fe$$

  1. Standard potentials cannot be added directly. We must use:

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

For the first reaction:

ΔG1∘=−(1)F(0.77)=−0.77F\Delta G_1^\circ = -(1)F(0.77) = -0.77FΔG1∘​=−(1)F(0.77)=−0.77F

For the second reaction:

ΔG2∘=−(2)F(−0.47)=+0.94F\Delta G_2^\circ = -(2)F(-0.47) = +0.94FΔG2∘​=−(2)F(−0.47)=+0.94F

  1. Add the free energy changes:

ΔGtotal∘=ΔG1∘+ΔG2∘\Delta G_{\text{total}}^\circ = \Delta G_1^\circ + \Delta G_2^\circΔGtotal∘​=ΔG1∘​+ΔG2∘​

ΔGtotal∘=−0.77F+0.94F=0.17F\Delta G_{\text{total}}^\circ = -0.77F + 0.94F = 0.17FΔGtotal∘​=−0.77F+0.94F=0.17F

  1. For the overall reaction, n=3n=3n=3 electrons:

ΔGtotal∘=−3FEoverall∘\Delta G_{\text{total}}^\circ = -3F E_{\text{overall}}^\circΔGtotal∘​=−3FEoverall∘​

So,

0.17F=−3FEoverall∘0.17F = -3F E_{\text{overall}}^\circ0.17F=−3FEoverall∘​

Eoverall∘=−0.173=−0.0567 VE_{\text{overall}}^\circ = -\frac{0.17}{3} = -0.0567\text{ V}Eoverall∘​=−30.17​=−0.0567 V

Eoverall∘≈−0.057 VE_{\text{overall}}^\circ \approx -0.057\text{ V}Eoverall∘​≈−0.057 V

  1. Therefore, the correct option is:

A: −0.057 V\boxed{\text{A: } -0.057\text{ V}}A: −0.057 V​

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