Question

Electrode potential of cadmium is $+0.4$V and electrode potential of chromium $=0.74$V $\left[C{d}^{2+}\right]=0.1$M and $\left[C{r}^{3+}\right]=0.01$M. Calculate the $\Delta G^o$ at $298$K.

Answer 1

$E_{cell}^o=E_{C{d}^{2+}|Cd}^o-E_{C{r}^{3+}|Cr}^o$

$=-0.4-(-0.74)=0.34V$

Total no. of electron exchanged, $n=6$

$\therefore \Delta G^o=-nF{E}_{cell}^o=-6\times 96500\times 0.34J$

$=-196860J$ .