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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

1 Answer

3 votes

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

User Holger Jakobs
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