Question

At temperature of $298K$ the emf of the
following electrochemical cell :
${\left(Ag\right)}_{(s)}\left|{\left(Ag\right)}^{+}(0.1M)\parallel {\left(Zn\right)}^{2+}(0.1M)\right|{\left(Zn\right)}_{(s)}$
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)

${\left(Ag\right)}_{(s)}\left|{\left(Ag\right)}_{(0.1m}\parallel {\left({\left(Zn\right)}^{2+}\right)}_{(0.1M)}\right|{\left(Zn\right)}_{(s)}$
$2Ag\to 2A{g}^{+}+2e^{-}$
$\bar{Z}{n}^{+2}+2e^{-}\to Zn$
$2{\left(Ag\right)}_{(s)}+{\left(Zn\right)}_{(aq)}\to 2{\left(Ag\right)}_{(zq)}+{\left(Zn\right)}_{(s)}$
$E_{\text{cell }}=-1.562-\frac{0.0591}{2}log\left(\frac{{\left(\left[{\left(Ag\right)}^{+}\right]\right)}^2}{\left[{\left(Zn\right)}^{+2}\right]}\right)$
$E_{cell}=-1.532V$