Question

A voltaic cell is constructed with two silver-silver chloride electrodes, where the halfreaction is AgCl (s) + e- → Ag (s) + Cl- (aq) E° = +0.222 V

The concentrations of chloride ion in the two compartments are 0.0222 M and 2.22 M, respectively. The cell emf is __________ V.

A) 0.212

B) 0.118

C) 0.00222

D) 22.2

E) 0.232

Answer 1

Answer: The cell potential of the cell is +0.118 V

Step-by-step explanation:

The half reactions for the cell is:

Oxidation half reaction (anode):
`Ag(s)+Cl^-(aq.)\rightarrow AgCl(s)+e^-`

Reduction half reaction (cathode):
`AgCl(s)+e^-\rightarrow Ag(s)+Cl^-(aq.)`

In this case, the cathode and anode both are same. So,
`E^o_(cell)` will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

`E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Cl^(-)]_(diluted))/([Cl^(-)]_(concentrated))`

where,

n = number of electrons in oxidation-reduction reaction = 1

`E_(cell)` = ?

`[Cl^(-)]_(diluted)` = 0.0222 M

`[Cl^(-)]_(concentrated)` = 2.22 M

Putting values in above equation, we get:

`E_(cell)=0-(0.0592)/(1)\log (0.0222M)/(2.22M)`

`E_(cell)=0.118V`

Hence, the cell potential of the cell is +0.118 V