Question

Electra - "Hmm, this looks tricky"

At ${25}^{\circ }C$,
$C{r}^{3+}+e^{-}\to C{r}^{2+},given{E}^0=-0.424VC{r}^{2+}+2e^{-}\to Cr\left(s\right),given{E}_0=-0.9VFind{E}^{0}at{25}^{\circ }CforC{r}^{3+}\left(aq\right)+3e^{-}\to Cr\left(s\right)$

Electra was completely stumped. Seems like you need a way to add electrode potentials. Please tell her the answer.

• A. 0.741 V
• B. - 0.741 V
• C. -1.324 V
• D. -0.476 V

Answer 1

The correct option is B - 0.741 V

We cannot add the electrode potentials directly because they are an intensive property. Think about density and temperature for example. You cannot just add the density of a rock and a paper directly to get the final density.

Gibb’s energy on the other hand, is an extensive property. We can add it directly.

Adding the first two reactions, we get the third equation and using the free energy concept, we have

$\Delta G_1^0+\Delta G_2^0=\Delta G_3^0$;

$-n_1{E}_1^{0}F-n_2{E}_2^{0}F=-n_3{E}_3^{0}F$
(n = number of electrons involved)

$E_3^0=\frac{n_1{E}_1^0+n_2{E}_2^0}{n_3}=\frac{1\times (-0.424)+2\times (-0.900)}{3}=-0.741V$