Question

Zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95℅ dissociated at this dilution at 298K.calculate the electrode Potential given that E¢ ( Zn2+|Zn) = - 0.76 V.​

Answer 1

Answer:

The electrode reaction is :

Zn2+(aq)+2e−→Zn(s)

According to Nernst equation :

E=E∘−0.0591nlog1[Mn+(aq)]

E=E∘−0.05912log1[Zn2+]

Since 0.1 M ZnSO4 solution is 95% dissociated, [Zn2+]=0.95×0.1=0.095M.

E=−0.76−0.05912log1(0.095)

=−0.76+0.02955 log 0.095=−0.76−0.02955×1.0223

=−0.76−0.0302=−0.7902V.

Explanation: