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Electrochemistry question

2023 · 30 Jan · Shift 1 · Q18
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  5. /2023 · 30 Jan · Shift 1 · Q18

Electrochemistry question

2023 · 30 Jan · Shift 1 · Q18

JEE MainChemistryElectrochemistryNumerical+4 / −1
Consider the cell Pt(s)∣H2( g,1 atm)∣H+(aq,1M)∣∣Fe3+(aq),Fe2+(aq)∣Pt⁡(s)\mathrm{Pt}_{(\mathrm{s})}\left|\mathrm{H}_{2}(\mathrm{~g}, 1 \mathrm{~atm})\right| \mathrm{H}^{+}(\mathrm{aq}, 1 \mathrm{M})|| \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq}) \mid \operatorname{Pt}(\mathrm{s})Pt(s)​∣H2​( g,1 atm)∣H+(aq,1M)∣∣Fe3+(aq),Fe2+(aq)∣Pt(s) When the potential of the cell is 0.712 V0.712 \mathrm{~V}0.712 V at 298 K298 \mathrm{~K}298 K, the ratio [Fe2+]/[Fe3+]\left[\mathrm{Fe}^{2+}\right] /\left[\mathrm{Fe}^{3+}\right][Fe2+]/[Fe3+] is ‾\underline{\hspace{2cm}}​. (Nearest integer) Given : Fe3++e−=Fe2+,EθFe3+,Fe2+∣Pt=0.7712.303RTF=0.06 V\mathrm{Fe}^{3+}+\mathrm{e}^{-}=\mathrm{Fe}^{2+}, \mathrm{E}^{\theta} \mathrm{Fe}^{3+}, \mathrm{Fe}^{2+} \mid \mathrm{Pt}=0.771 \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}Fe3++e−=Fe2+,EθFe3+,Fe2+∣Pt=0.771F2.303RT​=0.06 V
Numerical answer
View written solutionFree

Correct answer: 10

  1. Identify the half-cells

The cell is

Pt∣H2(1 atm)∣H+(1 M)∣∣Fe3+,Fe2+∣Pt\mathrm{Pt}|\mathrm{H_2}(1\,\text{atm})|\mathrm{H^+}(1\,\text{M})||\mathrm{Fe^{3+}},\mathrm{Fe^{2+}}|\mathrm{Pt}Pt∣H2​(1atm)∣H+(1M)∣∣Fe3+,Fe2+∣Pt

Left electrode is the standard hydrogen electrode (SHE), so

Eleft=0 VE_{\text{left}}=0\text{ V}Eleft​=0 V

Right electrode is

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

with

E∘=0.771 VE^\circ=0.771\text{ V}E∘=0.771 V

Since SHE is on the left and iron half-cell is on the right,

Ecell=Eright−Eleft=ErightE_{\text{cell}}=E_{\text{right}}-E_{\text{left}}=E_{\text{right}}Ecell​=Eright​−Eleft​=Eright​

So,

Eright=0.712 VE_{\text{right}}=0.712\text{ V}Eright​=0.712 V
  1. Apply the Nernst equation to the iron electrode

For the reduction

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

Nernst equation is

E=E∘−0.061log⁡[Fe2+][Fe3+]E=E^\circ-\frac{0.06}{1}\log\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}E=E∘−10.06​log[Fe3+][Fe2+]​

Substitute the values:

0.712=0.771−0.06log⁡[Fe2+][Fe3+]0.712=0.771-0.06\log\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}0.712=0.771−0.06log[Fe3+][Fe2+]​
  1. Solve for the ratio

Rearranging,

0.06log⁡[Fe2+][Fe3+]=0.771−0.712=0.0590.06\log\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}=0.771-0.712=0.0590.06log[Fe3+][Fe2+]​=0.771−0.712=0.059

Thus,

log⁡[Fe2+][Fe3+]=0.0590.06≈0.9833\log\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}=\frac{0.059}{0.06}\approx 0.9833log[Fe3+][Fe2+]​=0.060.059​≈0.9833

Therefore,

[Fe2+][Fe3+]=100.9833≈9.62\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}=10^{0.9833}\approx 9.62[Fe3+][Fe2+]​=100.9833≈9.62

Nearest integer:

10\boxed{10}10​
  1. Comparison with stored answer

Stored correct answer = 101010

Our derived answer = 101010

Hence, they agree.

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