Question

Electrode potential of cadmium is $-0.4$V and electrode potential of chromium is $-0.74$V. $\left[C{d}^{2+}\right]=0.1$M and $\left[C{r}^{3+}\right]=0.01$M. Calculate the $E_{cell}$ and $E_{cell}^o$.

Answer 1

$E_{cell}^o=\text{Electrode potential of cadmium - electrode potential of chromium}$

$E_{cell}^o=-0.4V-(-0.74V)=0.34V$

Now,

$\left[C{d}^{2+}\right]=0.1M,\left[C{r}^{3+}\right]=0.01M$

The cell reaction is:

$3C{d}^{2+}+2Cr\to 3Cd+2C{r}^{3+}$

Here, total no. of electrons exchanged are, $n=6$

$E_{cell}=E_{cell}^o+\frac{0.059}{n}\log \frac{[C{d}^{2+}{]}^3}{[C{r}^{3+}{]}^2}$

$=+0.34+\frac{0.059}{6}\log \frac{{0.1}^3}{{0.01}^2}=0.35V$.

$E_{cell}=0.35V$