Question

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

• 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
$2Ag(s)+Z{n}^{2+}(0.1M)⟶$
$2A{g}^{+}(0.1M)+Zn(s)$
The Nernst equation is
$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.0591}{n}\log \frac{{\left[A{g}^{+}\right]}^2}{\left[Z{n}^{2+}\right]}$
or, $E_{\text{cell }}=(-1.562)-\frac{(0.0591)}{2}\log \frac{(0.1{)}^2}{(0.1)}$
$($ where, $E_{\text{cell }}^{\circ }=-1.562V)$
or, $E_{\text{cell }}=(-1.562)-\frac{0.0591}{2}\log 10^{-1}$
$=-1.562+\frac{0.0591}{2}$
$=-1562+0.02955$
$=-1532V$