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CBSE Class 12 Chemistry 2015 Solved Paper

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Question : 21 of 26
Marks: +1, -0
Calculate emf of the following cell at 25∘C25^{\circ} \mathrm{C} :
  Fe∣Fe2+(0.001M)∥H+(0.01M)∣\; \mathrm{Fe} | \mathrm{Fe}^{2+}(0.001 \mathrm{M}) \parallel \mathrm{H}^{+}(0.01 \mathrm{M}) | H2(g)(bo1  bar  )∣Pt(s)\mathrm{H}_2(\mathrm{g}) (\text{bo} 1 \; \text{bar} \;) | \mathrm{Pt}(\mathrm{s})
  E∘(Fe2+∣Fe)=−0.44VE∘\; E^{\circ} (\mathrm{Fe}^{2+} | \mathrm{Fe}) = -0.44 \mathrm{V} E^{\circ} (H+∣H2)=0.00V(\mathrm{H}^{+} | \mathrm{H}_2) = 0.00 \mathrm{V}
Solution:  
The cell reaction:
  Fe(s)+2H+(aq)→Fe2+(aq)\; \mathrm{Fe}(\mathrm{s}) + 2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq}) +H2(g)+ \mathrm{H}_2(\mathrm{g})
  Ecell∘=Ec∘−Ea∘\; E^{\circ}_{\text{cell}} = E_{\mathrm{c}}^{\circ} - E_{\mathrm{a}}^{\circ}
    =[0−(−0.44)]V=0.44V\; \; = [0 - (-0.44)] \mathrm{V} = 0.44 \mathrm{V}
  Ecell=Ecell∘−0.0592log⁡[Fe2+][H+]2\; E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{2} \log \frac{ [\mathrm{Fe}^{2+}] }{ [\mathrm{H}^{+}]^2 }
  Ecell=0.44V−0.0592log⁡(0.001)(0.01)2\; E_{\text{cell}} = 0.44 \mathrm{V} - \frac{0.059}{2} \log \frac{(0.001)}{(0.01)^2}
    =0.44V−0.0592log⁡(10)\; \; = 0.44 \mathrm{V} - \frac{0.059}{2} \log (10)
  =0.44V−0.0295V\; = 0.44 \mathrm{V} - 0.0295 \mathrm{V}
    ≈0.410V\; \; \approx 0.410 \mathrm{V}
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