Question

An electrochemical cell has the following overall reaction:

Mn(s) + Cu2+(aq) -----> Cu(s) + Mn2+(aq)

The reduction half-reaction has a standard potential of 1.18 V.
The oxidation half-reaction has a standard potential of 0.34 V.
What is the overall cell potential?

(A) -1.52 V
(B) -1.18 V
(C) +1.18 V
(D) +1.52 V

Answer 1

Oxidation is a process in which electrons are lost from the substance. Since electrons are lost, oxidation state increases.

In the above reaction, we can see that oxidation state increases for Mn from 0 to +2. That mean Mn is undergoing oxidation.

`Mn(s)----->Mn^(2+)(aq) +2e^(-)`

We have
`E_(ox) = 0.34V`

The oxidation state decreases for Cu, that means Cu undergoes reduction.

`Cu^(2+)(aq) + 2e^(-) ------> Cu(s)`

`E_(red) =1.18V`

The formula to calculate cell potential is

`E_(cell) = E_(ox) + E_(red)`

Let us plug in the given Eox and Ered values.

`E_(cell) = 0.34V+1.18V`

`E_(cell) = 1.52V`

The overall cell potential is +1.52V