11
$\begingroup$

Given:

\begin{align} \ce{Co^{3+}(aq) + e- &-> Co^{2+}(aq)} & E° &= \pu{+1.82 V} \\ \ce{Co^{2+}(aq) + 2e- &-> Co(s)} & E° &= \pu{-0.28 V} \end{align}

what is the standard reduction potential for $\ce{Co^{3+}(aq) +3e- -> Co(s)}$?

If I add up these two half-equations to get the desired half-equation, I get $E^\circ = \pu{+1.54 V}$. But the correct answer is $\pu{+ 0.42 V}$!

Am I missing something?

$\endgroup$
14
$\begingroup$

That is because you cannot simply add the electrode potentials algebraically. In both cases, the electrode potentials have to be multiplied by $n$. What you can do instead is use Gibbs free energy change for each reaction which then can be added algebraically.

$$\Delta G^\circ = -nFE^\circ$$

Using this you can get the $\Delta G^\circ$ for each reaction which can be then added algebraically in the same way you add the reactions. The resultant $\Delta G^\circ_\mathrm{net}$, on equating with $-nFE^\circ_\mathrm{net}$, we can get the value of $E^\circ_\mathrm{net}$, which should be your answer.

In this approach, since the value of $n$ for both the reactions are different, the relative weightage of the electrode potentials of each reaction in determining the resultant $E^\circ$ will differ and therefore the final $E^\circ$ will be different than what you get by just adding up the electrode potentials.

For your case:

$$\begin{align} \ce{Co^3+ + e- &-> Co^2+} & E^\circ_1 &= \pu{+1.82 V} & \Delta G_1^\circ &= \pu{-175.6 kJ mol-1} \tag{1} \\ \ce{Co^2+ + 2e- &-> Co} & E^\circ_2 &= \pu{-0.28 V} & \Delta G_2^\circ &= \pu{+54.0 kJ mol-1} \tag{2} \\ \end{align}$$

$$\begin{align} \Delta G^\circ_\mathrm{net} &= \Delta G_1^\circ + \Delta G_2^\circ \\ &= \pu{-121.6 kJ mol-1} \\[3pt] E^\circ_\mathrm{net} &= -\frac{\pu{-121.6 kJ mol-1}}{3F} \\ &= \pu{+0.42 V} \end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.