Write the cell reaction and calculate the standard `E^(0)` of the cell: `Zn | Zn^(2+)(1M) | | Cd^(2+)(1M) | Cd `
Given `E_(Zn|Zn^(2+))^(0)= 0.763 `volt`E_(Cd|Cd^(2+))^(0) = 0.403` volt

Explanation

Reverse the signs of oxidation potential to get the values of the reduction or electrode potentials. Thus,`E_(Zn|Zn^(2+))^(0)=0.763`volt`:.` `E^(0) = - 0.763 ` and `rArr E_(Cd|Cd^(2+))^(0) = 0.403`
`:. E_(Cd^(2+)|Cd)^(0)` = -0.403V
Since Zn2+|Zn electrode is at lower potential, therefore, it acts as the anode while Cd2+|Cd electrode with higher potential acts as the cathode.
In other words, Zn loses electrons and `Cd^(2+)` ion accepts them. Therefore, cell reaction is :
`Zn + Cd^(2+) rarr Zn^(2+) + Cd`
And `E_(cell)^(0) = E_(Cd^(2+)| Cd)^(0) – E_(Zn^(2+)| Zn)^(0)` = -0.403 – (- 0.763) = +0.360 V