Question

Why is `Delta G` negative for electrolysis reactions?

Answer 1

`DeltaG^@>0` but after applying a potential `E_(cell)>=2.06V` from an external power source, `DeltaG` 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: `2H_2O->O_2+4H^(+)+4e^(-)" " "-E^@=-1.23V`

Cathode: `4H_2O+4e^(-)->2H_2+4OH^-" "E^@=-0.83V`

Net reaction: `6H_2O->2H_2+O_2+underbrace(4(H^(+)+OH^-))_(4H_2O)`

`2H_2O->2H_2+O_2" "E_(cell)^@=-2.06V`

A negative cell potential implies non spontaneous process and therefore, `DeltaG^@>0`.

Note that the relationship between `DeltaG^@` and `E^@` is given by:

`DeltaG^@=-nFE^@`

where, `n` is the number of electrons transferred during redox, which is `n=4` in this case,
and `F=96485C/("mol "e^-)` is Faraday's constant.

Therefore, since `E^@<0` `=>DeltaG^@>0`

Because `DeltaG^@>0`, thus after applying a potential `E_(cell)>=2.06V` from an external power source, `DeltaG` becomes negative and the reaction will be spontaneous.

Note that, `DeltaG=-nFE`

Electrochemistry | Electrolysis, Electrolytic Cell & Electroplating.