Why is #Delta G# negative for electrolysis reactions?

1 Answer
Dr. Hayek
Jan 28, 2016

#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.

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