# Cell Potential Formula

The driving force of the electron flow from anode to cathode shows a potential drop in the energy of the electrons moving into the wire.  The difference in potential energy between the anode and cathode is known as the cell potential in a voltaic cell.

Formula for cell potential is

E∘cellcell∘ = E∘reductionreduction∘ + E∘oxidationoxidation∘

(or)

Ecellcell = 0.0591n0.0591n log10 C2C1C2C1.

Problem 1: A concentration cell was created by immersing two silver electrodes in 0.05M and 0.1M AgNO3 solution. Write cell representation, cell reactions and find the EMF of the concentration cell.

Answer:

The cell representation

Ag|AgNO3 (0.05M) || AgNO3 (0.1M)|Ag

At anode,

Ag →→ Ag+ (0.05M) + e

At cathode,

Ag+ (0.1M) + e →→ Ag

Net cell reaction is,

Ag+ (0.1M) →→ Ag+ (0.05M)

The EMF of a concentration cell,

Ecellcell = 0.0591n0.0591n log10 C2C1C2C1

Ecellcell = 0.059110.05911 log10 0.10.050.10.05

Ecellcell = 0.12 V

Question 2: If a concentration cell of the silver electrode is represented as,

Ag/Ag+ (0.00475 M) || Ag+ (0.043 M)/Ag.

An electrode of higher concentration having Ag+ (0.043 M) acts as cathode while the other silver electrode having Ag+ ion concentration (0.00475 M) acts as anode.

Solution:

At anode,

Ag(s) →→ Ag+ + e

At cathode,

Ag+ + e →→ Ag(s)

The net reaction of the concentration cell is,

Ag+ (0.043 M) →→ Ag+ (0.00475 M)

The EMF of the concentration cell is,

Ecellcell = 0.0591n0.0591n log10 C2C1C2C1

Ecellcell = 0.059120.05912 log10 0.0430.004750.0430.00475

Ecellcell = 0.056 V