Question

Calculate the emf of the cell in which of the following reaction takes place
$Ni (s)+2 Ag ^{+}(0.002 \,M ) \longrightarrow Ni ^{2+}(0.160 \,M ) + 2Ag(s)$
(Give that $ E_{\text {cell }}^{\circ}=1.05\, V )$

• A. 0.73 V
• B. 0.91 V
• C. 0.62 V
• D. 0.34 V

Answer 1

Answer: (B)

From the given cell reaction and Nernst equation,
$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{n} \log \frac{\left[ Ni ^{2+}\right]}{\left[ Ag ^{+}\right]^{2}} $
$=1.05\, V -\frac{0.0591}{2} \log \frac{[0.160]}{[0.002]^{2}} $
$=1.05-\frac{0.0591}{2} \log \left(4 \times 10^{4}\right)$
$=1.05-\frac{0.0591}{2}(4.6021) $
$=1.05-0.14=+0.91 \,V $
$E_{\text {cell }}=+0.91\, V$