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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

Respuesta :

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.

[tex] Mn(s)----->Mn^{2+}(aq) +2e^{-} [/tex]

We have [tex] E_{ox} = 0.34V [/tex]

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

[tex] Cu^{2+}(aq) + 2e^{-} ------> Cu(s) [/tex]

[tex] E_{red} =1.18V [/tex]

The formula to calculate cell potential is

[tex] E_{cell} = E_{ox} + E_{red} [/tex]

Let us plug in the given Eox and Ered values.

[tex] E_{cell} = 0.34V+1.18V [/tex]

[tex] E_{cell} = 1.52V [/tex]

The overall cell potential is +1.52V