Zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95℅ disso

1.	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» 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:

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»

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: