Question

At temperature of $298K$ the emf of the
following electrochemical cell :
$\left(Ag\right)_{\left(\right. s \left.\right)}\left|\left(Ag\right)^{+} \left(\right. 0 . 1 M \left.\right) \parallel \left(Zn\right)^{2 +} \left(\right. 0 . 1 M \left.\right)\right|\left(Zn\right)_{\left(\right. s \left.\right)}$
will be (given ) $E_{Ag + / Ag}=0.80,E^\circ zn2+/Zn=-0.76$

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

Answer 1

Answer: (A)

$\left(Ag\right)_{\left(\right. s \left.\right)}\left|\left(Ag\right)_{\left(\right. 0 . 1 m} \parallel \left(\left(Zn\right)^{2 +}\right)_{\left(\right. 0 . 1 M \left.\right)}\right|\left(Zn\right)_{\left(\right. s \left.\right)}$
$2Ag \rightarrow 2Ag^{+}+2e^{-}$
$\bar{Z}n^{+ 2}+2e^{-} \rightarrow Zn$
$2\left(Ag\right)_{\left(\right. s \left.\right)}+\left(Zn\right)_{\left(\right. aq \left.\right)} \rightarrow 2\left(Ag\right)_{\left(\right. zq \left.\right)}+\left(Zn\right)_{\left(\right. s \left.\right)}$
$E_{\text{cell }}=-1.562-\frac{0 . 0591}{2}log\left(\frac{\left(\left[\left(Ag\right)^{+}\right]\right)^{2}}{\left[\left(Zn\right)^{+ 2}\right]}\right)$
$E_{cell}=-1.532V$