Explanation:
The given reaction at cathode will be as follows.
At cathode:
,
= -0.761 V
At anode:
,
= 0.761
Therefore, net reaction equation will be as follows.

Initial: 0.129 - - 0.427
Change: -0.047 - - -0.047
Equilibrium: (0.129 - 0.047) (0.427 - 0.047)
= 0.082 = 0.38
As
for the given reaction is zero.
Hence, equation for calculating new cell potential will be as follows.
E_{cell} = ![E^{o}_{cell} - \frac{RT}{nF} ln \frac{[Zn^{2+}]_{products}}{[Zn^{2+}]_{reactants}}](https://tex.z-dn.net/?f=E%5E%7Bo%7D_%7Bcell%7D%20-%20%5Cfrac%7BRT%7D%7BnF%7D%20ln%20%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D_%7Bproducts%7D%7D%7B%5BZn%5E%7B2%2B%7D%5D_%7Breactants%7D%7D)
= 
= 0.019
Thus, we can conclude that the cell potential of the given cell is 0.019.