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NCERT Class XII Chemistry
Chapter - Electrochemistry
Questions with Solutions

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Question : 4 of 32
Marks: +1, -0
Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
(i)2Cr(s)+3Cd(aq)2+→2Cr(aq)3++3Cd(s)2 \mathrm{Cr}_{(s)} + 3 \mathrm{Cd}^{2+}_{(aq)} \rightarrow 2 \mathrm{Cr}^{3+}_{(aq)} + 3 \mathrm{Cd}_{(s)}
(ii) Fe(aq)2++Ag(aq)+→Fe(aq)3++Ag(s)\mathrm{Fe}^{2+}_{(aq)} + \mathrm{Ag}^{+}_{(aq)} \rightarrow \mathrm{Fe}^{3+}_{(aq)} + \mathrm{Ag}_{(s)}
Calculate the ΔrG∘\Delta_r G^\circ and equilibrium constant of the reactions.
Given : ECr3+/Cr∘=−0.74 V,ECd2+/Cd∘=−0.40 VE^\circ_{\mathrm{Cr}^{3+}/\mathrm{Cr}} = -0.74\,\text{V}, \quad E^\circ_{\mathrm{Cd}^{2+}/\mathrm{Cd}} = -0.40\,\text{V}
EAg+/Ag∘=0.80 V,EFe3+/Fe2+∘=0.77 VE^\circ_{\mathrm{Ag}^{+}/\mathrm{Ag}} = 0.80\,\text{V}, \quad E^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = 0.77\,\text{V}
Solution:  
(i) Ecell∘=Ecathode∘−Eanode∘E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}} =−0.40 V−(−0.74 V)=+0.34 V= -0.40\,\text{V} - (-0.74\,\text{V}) = +0.34\,\text{V}
ΔG∘=−nFEcell∘\Delta G^\circ = -n F E^\circ_{\mathrm{cell}} =−6×96500×0.34=−196860 J mol−1= -6 \times 96500 \times 0.34 = -196860\,\mathrm{J\,mol}^{-1} =−196.86 kJ mol−1= -196.86\,\mathrm{kJ\,mol}^{-1}
−ΔG∘=2.303RTlog⁡K-\Delta G^\circ = 2.303 R T \log K
196860=2.303×8.314×298×log⁡K196860 = 2.303 \times 8.314 \times 298 \times \log K or log⁡K=34.5014\log K = 34.5014
K=Antilog 34.5014K = \text{Antilog}\,34.5014 =3.172×1034= 3.172 \times 10^{34}
(ii) E°cell_{\mathrm{cell}} = +0.80 V – 0.77 V = +0.03 V
ΔG° = –nFE°cell_{\mathrm{cell}} = – 1 × 96500 × 0.03 = –2895 J mol−1\mathrm{mol}^{-1} = –2.895 kJ mol−1\mathrm{mol}^{-1}
ΔG°= –2.303 RT log K
–2895 = –2.303 × 8.314 × 298 × log K
or log K = 0.5074
K = Antilog (0.5074) = 3.22
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