Question

For the redox reaction
$ \ce{Zn(s) +Cu^{2+} (0.1\,M) -> Zn^{2+} (1\,M) +Cu(s)} $ taking place in a cell, $ E^{\circ}_{cell} $ is $ 1.10\,V $ . $ E_{cell} $ for the cell will be $ \left(2.303 \frac{RT}{F}=0.0591\right) $

• A. $ 2.41\,V $
• B. $ 1.80\,V $
• C. $ 1.07\,V $
• D. $ 0.82\,V $

Answer 1

Answer: (C)

$E_{cell}=E^{\circ}_{cell}-\frac{0.0591}{n} log\,Q$
$\underset{\text{0.1 M}}{\ce{Cu^{2+}}} + \ce{Zn -> }$ $\underset{\text{1 M}}{\ce{Zn^{2+}}} + \ce{Cu}$
$Q=\frac{\left[Zn^{2+}\right]}{\left[Cu^{2+}\right]}=\frac{1}{0.1}=10$
$E_{cell}=1.10-\frac{0.0591}{2} log\,10$
$= 1.10 − 0.0295$
$= 1.0705 \,V$