-->

Electrochemistry

Question
CBSEENCH12006005

Zn | Zn2+ (α = 0.1 M) || Fe2+ (α = 0.01 M) | Fe. The emf of the above cell is 0.2905 V. What is the equilibrium constant for the cell reaction?

Solution

For cell
Zn|Zn2+(α = 0.1 M) || Fe2+ (α = 0.01 M) | Fe
The cell reaction
 (i) Zn(s)  Zn2+(aq) + 2e(ii) Fe2+(aq) + 2e  Fe(s)

The overall reaction isZn(s) + Fe2+(aq) Zn2+(aq)  +Fe(s)

On applying Nernst equation

                    Ecell = E°cell - 0.0591nlog 10Zn2+Fe2+0.2905 = E°cell-0.05912log100.10.01


or              0.2905 = E°cell-0.0295×log1010

or              0.2905  = E°cell - 0.0295 × 1

or                E°cell = 0.2905 + 0.0295 = 0.32 V
   

At equilibrium (Ecell = 0)

                    Ecell = E°cell - 0.0591nlog10Kc

∴         0 = E°cell - 0.0591nlog10 Kc

or      E°cell = 0.0591nlog10 Kc

or      0.32 = 0.0591n log10 Kc

or               Kc = 100.32/0.0295.