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NCERT Class XII Chemistry
Chapter - Electrochemistry
Questions with Solutions

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Question : 23 of 32
Marks: +1, -0
Calculate the emf of the cell in which the following reaction takes place:
Ni(s)+2Ag+(0.002 M)→Ni2+(0.160 M)+2Ag(s)\mathrm{Ni}_{(s)} + 2\mathrm{Ag}^{+}(0.002\ \text{M}) \rightarrow \mathrm{Ni}^{2+}(0.160\ \text{M}) + 2\mathrm{Ag}_{(s)}
Given thatEcell∘E^{\circ}_{\text{cell}}= 1.05 V
Solution:  
Ni(s)→Ni(aq)2++2e−\mathrm{Ni}_{(s)} \rightarrow \mathrm{Ni}^{2+}_{(aq)} + 2e^{-}
2Ag(aq)++2e−→2Ag(s)2\mathrm{Ag}^{+}_{(aq)} + 2e^{-} \rightarrow 2\mathrm{Ag}_{(s)}
—\text{---}
Ni(s)+2Ag(aq)+→Ni(aq)2++2Ag(s)\mathrm{Ni}_{(s)} + 2\mathrm{Ag}^{+}_{(aq)} \rightarrow \mathrm{Ni}^{2+}_{(aq)} + 2\mathrm{Ag}_{(s)}
—\text{---}
Ecell∘=Ecell∘−0.05912log⁡[Ni2+][Ag+]2E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{2} \log \frac{[\mathrm{Ni}^{2+}]}{[\mathrm{Ag}^{+}]^{2}}
=1.05−0.05912log⁡0.16(0.002)2= 1.05 - \frac{0.0591}{2} \log \frac{0.16}{(0.002)^{2}}
=1.05−0.05912log⁡0.164×10−6= 1.05 - \frac{0.0591}{2} \log \frac{0.16}{4 \times 10^{-6}}
=1.05−0.05912log⁡4×104= 1.05 - \frac{0.0591}{2} \log 4 \times 10^{4}
=1.05−0.05912(log⁡4+log⁡104)= 1.05 - \frac{0.0591}{2} (\log 4 + \log 10^{4})
=1.05−0.05912(0.602+4.0000)= 1.05 - \frac{0.0591}{2}(0.602+4.0000)
=1.05−0.0591×4.6022= 1.05 - \frac{0.0591 \times 4.602}{2}
=1.05−0.1358=0.9142= 1.05 - 0.1358 = 0.9142
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