56. Zinc rod is dipped in 0.1M solution of ZnSO4. The salt is 95% dissociated at this dilution at 298K Calculateelectrode potential. (E°Zn2+/Zn=-0.76V) [Ans: -0.7902V]
Answers
Answered by
4
Answer:
Electrode potential E = -0.7902V.
Expiation:
Let us understand the Electrode Reaction:
Zn2 + (aq) + 2e− → Zn(s)Zn2 + (aq) + 2e- → Zn(s)
As Per The Nernst Equation:
E = E∘ − 0.0591nlog1[Mn+(aq)]E = E∘-0.0591nlog1[Mn+(aq)]
E = E∘−0.05912log1[Zn2+]E = E∘-0.05912log1[Zn2+]
It is given that,
0.1 M ZnSO4ZnSO4 solution is 95%
Disassociated at This Dilution,
[Zn2+] = 0.95×0.1
= 0.095M[Zn2+]
= 0.95×0.1
= 0.095M.
E = −0.76−0.05912log1(0.095)E
= -0.76-0.05912log1(0.095)
= −0.76+0.02955 log 0.095
= −0.76−0.02955×1.0223
= -0.76+0.02955 log 0.095
= -0.76-0.02955×1.0223
= −0.76−0.0302
= −0.7902V
= -0.76-0.0302
E = -0.7902V.
Similar questions