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CBSE Class 12 Chemistry 2013 Solved Paper

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Question : 21 of 30
Marks: +1, -0
Calculate the emf of the following cell at 298 K298\,\text{K} :
  Fe(s)∣Fe2+(0.001 M)∥H+(1 M)∣H2(g)(1 bar),\;\mathrm{Fe}(s) \vert \mathrm{Fe}^{2+}(0.001\,\text{M}) \Vert \mathrm{H}^{+}(1\,\text{M}) \vert \mathrm{H}_2(g)(1\,\text{bar}),
  Pt(s)\;\mathrm{Pt}(s)
  (  Given  E∘  cell  =+0.44 V)\;\left( \;\text{Given}\; {E^{\circ}}_{\;\text{cell}\;}=+0.44\,\text{V} \right)
Solution:  
According to given equation:
Fe(s)+2 H+(aq)→Fe+2(aq)+H2(g)\mathrm{Fe}(s) + 2\,\mathrm{H}^{+}(aq) \rightarrow \mathrm{Fe}^{+2}(aq) + \mathrm{H}_2(g)
E∘cell=0.44 V{E^{\circ}}_{\text{cell}} = 0.44\,\text{V}
By applying Nernst equation:
Ecell=Ecell∘−0.0591nlog⁡[Fe+2][H+]2E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\mathrm{Fe}^{+2}]}{[\mathrm{H}^{+}]^2}
Ecell=0.44−0.05912log⁡0.001(1)2E_{\text{cell}} = 0.44 - \frac{0.0591}{2} \log \frac{0.001}{(1)^2}
Ecell=0.44−0.0295log⁡10−3E_{\text{cell}} = 0.44 - 0.0295 \log 10^{-3}
Ecell=0.44−0.0295(−3log⁡10)E_{\text{cell}} = 0.44 - 0.0295(-3 \log 10)
where [log⁡10=1][\log 10 = 1]
=0.44−0.0295(−3×1)= 0.44 - 0.0295(-3 \times 1)
=0.44+0.0885= 0.44 + 0.0885
Ecell=0.528 VE_{\text{cell}} = 0.528\,\text{V}
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