Question

Calculate the emf of the call in which of the following reaction:
$Ni\left(s\right)+2A{g}^{+}\left(0.002M\right)\to N{i}^{2+}\left(0.160M\right)+2Ag\left(s\right)$

(Give that $E_{cell}^o=1.05V$)

• A. 0.3 V
• B. 0.91 V
• C. 0.62 V
• D. 0.34 V

Answer 1

The correct option is B 0.91 V

$\left[A{g}^{+}\right]=0.002M$
$\left[N{i}^{2+}\right]=0.160M$

According to Nernst equation:

$E_{cell}=E^0-\frac{0.0591}{n}log\frac{\left[N{i}^{2+}\right]}{[A{g}^{+}{]}^2}$

$E_{cell}=1.05-\frac{0.0591}{2}log\frac{0.160}{{0.002}^2}$

$E_{cell}=1.05-0.02955\times log(4\times {10}^4)$

$E_{cell}=1.05-0.02955\times \left(4.6021\right)$

$E_{cell}=0.914V$

Hence option $B$ is correct.