Question

At temperature of $298\, K$ the emf of the following electrochemical cell
$Ag\left(s\right) | Ag^{+} \left(0.1\, M\right) | |Zn^{2+}\left(0.1\, M\right) | Zn\left(s\right)$ will be (given $E^{o}_{cell} = - 1.562\, V$)

• A. - 1.532 V
• B. - 1.503 V
• C. 1.532 V
• D. - 3.06 V

Answer 1

Answer: (A)

From the given cell, the cell reaction is
$2 Ag ( s )+ Zn ^{2+}(0.1\, M ) \longrightarrow $
$2 Ag ^{+}(0.1\, M )+ Zn (s)$
The Nernst equation is
$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{n} \log \frac{\left[ Ag ^{+}\right]^{2}}{\left[ Zn ^{2+}\right]}$
or, $E_{\text {cell }}=(-1.562)-\frac{(0.0591)}{2} \log \frac{(0.1)^{2}}{(0.1)}$
$\left(\right.$ where, $\left. E_{\text {cell }}^{\circ}=-1.562\, V \right)$
or, $E_{\text {cell }} =(-1.562)-\frac{0.0591}{2} \log 10^{-1}$
$=-1.562+\frac{0.0591}{2}$
$=-1562+0.02955$
$=-1532\, V$