# Why is Delta G negative for electrolysis reactions?

##### 1 Answer
Jan 28, 2016

$\Delta {G}^{\circ} > 0$ but after applying a potential ${E}_{c e l l} \ge 2.06 V$ from an external power source, $\Delta G$ becomes negative and the reaction will be spontaneous.

#### Explanation:

Let us discuss the example of electrolysis of water.

In electrolysis of water, hydrogen and oxygen gases are produced.

The anode and the cathode half-reactions are the following:

Anode: $2 {H}_{2} O \to {O}_{2} + 4 {H}^{+} + 4 {e}^{-} \text{ " } - {E}^{\circ} = - 1.23 V$

Cathode: $4 {H}_{2} O + 4 {e}^{-} \to 2 {H}_{2} + 4 O {H}^{-} \text{ } {E}^{\circ} = - 0.83 V$

Net reaction: $6 {H}_{2} O \to 2 {H}_{2} + {O}_{2} + {\underbrace{4 \left({H}^{+} + O {H}^{-}\right)}}_{4 {H}_{2} O}$

$2 {H}_{2} O \to 2 {H}_{2} + {O}_{2} \text{ } {E}_{c e l l}^{\circ} = - 2.06 V$

A negative cell potential implies non spontaneous process and therefore, $\Delta {G}^{\circ} > 0$.

Note that the relationship between $\Delta {G}^{\circ}$ and ${E}^{\circ}$ is given by:

$\Delta {G}^{\circ} = - n F {E}^{\circ}$

where, $n$ is the number of electrons transferred during redox, which is $n = 4$ in this case,
and $F = 96485 \frac{C}{\text{mol } {e}^{-}}$ is Faraday's constant.

Therefore, since ${E}^{\circ} < 0$ $\implies \Delta {G}^{\circ} > 0$

Because $\Delta {G}^{\circ} > 0$, thus after applying a potential ${E}_{c e l l} \ge 2.06 V$ from an external power source, $\Delta G$ becomes negative and the reaction will be spontaneous.

Note that, $\Delta G = - n F E$

Electrochemistry | Electrolysis, Electrolytic Cell & Electroplating.