Calculate the emf of the cell in which the following reaction takes place:
$\mathrm{Ni}_{(s)}+2 \mathrm{Ag}^{+}(0.002 \mathrm{M}) \rightarrow \mathrm{Ni}^{2+}(0.160 \mathrm{M})+2 \mathrm{Ag}_{(s)}$
Given that $E_{\text {(cell) }}^{\ominus}=1.05 \mathrm{~V}$
Applying Nernst equation we have:
$E_{\text {(cell) }}=E_{\text {(cell) }}^{\ominus}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Ni}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}$
$=1.05-\frac{0.0591}{2} \log \frac{(0.160)}{(0.002)^{2}}$
$=1.05-0.02955 \log \frac{0.16}{0.000004}$
= 1.05 − 0.02955 log 4 × 104
= 1.05 − 0.02955 (log 10000 + log 4)
= 1.05 − 0.02955 (4 + 0.6021)
= 0.914 V
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