# Electrolysis in aqueous solution – which equations to use to predict product at each electrode?

When trying to work out what will be produced at each electrode during electrolysis, I'm confused as to which equations to use for hydrogen and oxygen.

For instance, an electrolytic cell containing aqueous nickel(II) bromide, with a nickel cathode and a platinum anode.

At the cathode:

$$\ce{Ni^2+ + 2e- -> Ni}$$

I have read that this does not occur because $\ce{Ni^2+}$ is a worse oxidising agent than $\ce{H+}$, which makes sense because $\ce{H+}$ is lower down on the left side of the electrochemical series meaning its reduction is more feasible.

Water is therefore reduced at the cathode:

$$\ce{2H+ + 2e- -> H2}\quad E^\circ=0.0\ \mathrm V$$

or:

$$\ce{2H2O + 2e- -> H2 + 2OH-}\quad E^\circ=-0.83\ \mathrm V$$

The second equation, however, has water higher up in the electrochemical series. This would suggest nickel should actually be reduced in preference to water.

How do I know which of the two "water-reduction" equations to use?

This is even less transparent for the anode, where bromide should be oxidised to bromine, but there are also two equations showing production of oxygen: one below and one above bromide in the series:

\begin{alignat}{2} \ce{O2 + 4H+ + 4e- &-> 2H2O}\quad &E^\circ&=+1.23\ \mathrm V\\[6pt] \ce{O2 + 2H2O + 4e- &-> 4OH-}\quad &E^\circ&=+0.40\ \mathrm V \end{alignat}

I don't understand which of these equations should be used and how you're supposed to know which to use.

The given potentials $E = 0.00{\text{ V}}$ and $E = -0.83{\text{ V}}$ for the reduction of water apply to different $\text{pH}$.

The thermodynamic relation of the potential $E$ to the composition of the solution is generally known as the Nernst equation:

$$E = E^\circ + \frac{{0.059{\text{ V}}}}{z}\log \frac{{{{\prod\nolimits_i {\left[ {{\text{ox}}} \right]} }^{{n_i}}}}}{{{{\prod\nolimits_j {\left[ {{\text{red}}} \right]} }^{{n_j}}}}}$$

For the given reaction

$$\ce{2 H+ + 2e- <=> H2}$$

$$E = 0.00{\text{ V}} + \frac{{0.059{\text{ V}}}}{2}\log \frac{{{{\left[ {{{\text{H}}^ + }} \right]}^2}}}{{\left[{{{\text{H}}_2}}\right]}}$$

Since

$${\text{pH}} = - \log \left[ {{{\text{H}}^ + }} \right]$$

the potential $E$ depends on $\text{pH}$:

$$E = 0.00{\text{ V}} - 0.059{\text{ V}} \times {\text{pH}} + \frac{{0.059{\text{ V}}}}{2}\log \frac{1}{{\left[ {{{\text{H}}_2}} \right]}}$$

Therefore, the potential is $E = 0.00{\text{ V}}$ at $\text{pH} = 0$ and $E = -0.83{\text{ V}}$ at $\text{pH} = 14$ respectively.

• Thank you for this answer, though it is a little over my head. It still doesn't answer the question of which equation to use when trying to report what process is happening at each electrode. Which half-equation would be most suitable for explaining the production of oxygen or hydrogen at each electrode? – Raf Mar 4 '15 at 7:02
• @Raf Since the potential of water depends on pH, you may use your first equation at E = 0.00 V for an acidic solution and the second equation at E = −0.83 V for an alkaline solution, respectively. Or better still, you calculate the actual potential E for the actual pH of the solution (using the formula given in my answer). (The oxidation reaction is analogous.) – Loong Mar 4 '15 at 12:46
• Thank you Loong. I've realised what I was looking for was an answer to my question without using the Nernst equation, but it just can't be that simple. – Raf Mar 5 '15 at 5:20