Answer:
The cell voltage of the given cell is 2.01 V
Explanation:

Oxidation half reaction:

Reduction half reaction:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the
of the reaction, we use the equation:

To calculate the EMF of the cell, we use the Nernst equation, which is:
![E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B2.303RT%7D%7BnF%7D%5Clog%20%5Cfrac%7B%5BAl%5E%7B3%2B%7D%5D%5E2%7D%7B%5BCu%5E%7B2%2B%7D%5D%5E3%7D)
where,
= electrode potential of the cell = ? V
= standard electrode potential of the cell = +2.00 V
R = Gas constant = 8.314 J/mol.K
T = temperature =
=[42+273]K=315K
F = Faraday's constant = 96500
n = number of electrons exchanged = 6
![[Cu^{2+}]=3.43M[Al^{3+}]=1.63M](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D3.43M%5BAl%5E%7B3%2B%7D%5D%3D1.63M)


Thus, the cell voltage of the given cell is 2.01