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CBSE Class 12 Chemistry 2016 Delhi Set 1 Solved Paper

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Question : 16 of 26
Marks: +1, -0
Calculate e.m.f. of the following cell at 298 K298\,\text{K} :
  2 Cr(s)+3 Fe2+(0.1 M)→2 Cr3+(0.01 M)\;2\,\mathrm{Cr}(\mathrm{s}) + 3\,\mathrm{Fe}^{2+}(0.1\,\text{M}) \rightarrow 2\,\mathrm{Cr}^{3+}(0.01\,\text{M}) +3 Fe(s)+3\,\mathrm{Fe}(\mathrm{s})
    Given:  E(Cr3+∣Cr)∘=−0.74 V E∘\;\; \text{Given:}\; E^\circ_{(\mathrm{Cr}^{3+} \mid \mathrm{Cr})} = -0.74\ \text{V}\ E^\circ
E(Fe2+∣Fe)∘=−0.44 V 3E^\circ_{(\mathrm{Fe}^{2+} \mid \mathrm{Fe})} = -0.44\ \text{V}\ 3
Solution:  
E  celi  ∘=Ec0−Ea0E^\circ_{\;\text{celi}\;} = E_c^0 - E_a^0
=(−0.44)−(−0.74) V= (-0.44) - (-0.74)\ \text{V}
=0.30 V= 0.30\ \text{V}
E  cell  =E  cell  ∘−0.059nlog⁡[Cr3+]2[Fe2+]3E_{\;\text{cell}\;} = E^\circ_{\;\text{cell}\;} - \frac{0.059}{n} \log \frac{[\mathrm{Cr}^{3+}]^2}{[\mathrm{Fe}^{2+}]^3}
E  cell  =E  cell  ∘−0.0596log⁡[0.01]2[0.1]3E_{\;\text{cell}\;} = E^\circ_{\;\text{cell}\;} - \frac{0.059}{6} \log \frac{[0.01]^2}{[0.1]^3}
=0.30−(−0.0596)= 0.30 - \left(\frac{-0.059}{6}\right)
=0.3098 V= 0.3098\ \text{V}
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