Respuesta :
Explanation:
The given reaction at cathode will be as follows.
At cathode: [tex]Zn^{2+} + 2e^{-} \rightarrow Zn[/tex], [tex]E_{o}[/tex] = -0.761 V
At anode: [tex]Zn \rightarrow Zn^{2+} + 2e^{-}[/tex], [tex]E_{o}[/tex] = 0.761
Therefore, net reaction equation will be as follows.
[tex]Zn^{2+} + Zn \rightarrow Zn + Zn^{2+}[/tex]
Initial: 0.129 - - 0.427
Change: -0.047 - - -0.047
Equilibrium: (0.129 - 0.047) (0.427 - 0.047)
= 0.082 = 0.38
As [tex]E^{o}_{cell}[/tex] for the given reaction is zero.
Hence, equation for calculating new cell potential will be as follows.
E_{cell} = [tex]E^{o}_{cell} - \frac{RT}{nF} ln \frac{[Zn^{2+}]_{products}}{[Zn^{2+}]_{reactants}}[/tex]
= [tex]0 - \frac{8.314 atm L/mol K \times 291 K}{2 \times 96500} ln \frac{0.38}{0.082}[/tex]
= 0.019
Thus, we can conclude that the cell potential of the given cell is 0.019.
