Question

Calculate the emf of the cell in which the following reaction takes place: 

\[\ce{Ni_{(s)} + 2Ag+ (0.002 M) -> Ni{2+} (0.160 M) + 2Ag_{(s)}}\]

Given that \[\ce{E^\circ_{cell}}\] = 1.05 V

Answer 1

\[\ce{Ni_{(s)} -> Ni{^{2+}_{(aq)}} + 2e-}\]

\[\ce{2Ag{^+_{(aq)}} + 2e- -> 2Ag_{(s)}}\]
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\[\ce{Ni_{(s)} + 2Ag{^+_{(aq)}} -> Ni^{2+}_{(aq)} + 2Ag_{(s)}}\]

Applying the Nernst equation, we have:

E_cell = \[\ce{E^{\circ}_{cell} - \frac{0.0591}{2} log \frac{[Ni^{2+}]}{[Ag^+]^2}}\]

= \[\ce{1.05 - \frac{0.0591}{2} log \frac{0.16}{(0.002)^2}}\]

= \[\ce{1.05 - \frac{0.0591}{2} log \frac{0.16}{4 \times 10^{-6}}}\]

= \[\ce{1.05 - \frac{0.0591}{2} log 4 \times 10^4}\]

= \[\ce{1.05 - \frac{0.0591}{2} (log 4 + log 10^4)}\]

= \[\ce{1.05 - \frac{0.0591}{2} (0.602 + 4.0000)}\]

= \[\ce{1.05 - \frac{0.0591 \times 4.602}{2}}\]

= 1.05 − 0.1359

= 0.9142 V