How do i make this cell spontaneous?

Question

its a galvanic cell, with 2 half reactions... ihave to draw the cell blah blah but i first have to reverse 1 reaction in order to make it positive and spontaneous.
but im really confused, since reducation takes place at the cathode, and oxidation happens at the anode

which half reaction do i reverse?
Mn^+2 + 2e- --> Mn E*= -1.18V
Fe^+3 + 3e- -->Fe E*= -0.036

Ecell = Ecathode- E anode

im stuck!!!

Answer 1

To help you clarify things

Let's assume we have the general reaction

A(+2)+ B -> A + B(+2)

It comes from the half reactions

A(+2) +2e -> A with E1
B-> B(+2) + 2e with E2

and has Etotal = E1+E2
Now since in all the tables you get the E for reduction half-reactions, the value you will have for the second half-reaction is E'2 = -E2.
Thus Etotal =E1+(-E'2) = E1- E'2

In your case let's inverse the first reaction. Then

Mn -> Mn(+2) + 2e E'Mn= -EMn= +1.18

So for the total reaction you have
Ecell = E(Fe)+E'(Mn)= -0.036+1.18 = +1.144 V

Notes:
1. The complete reaction is
2Fe(+3) + 3 Mn -> 2Fe + 3Mn(+2)
The stoichiometric coefficients are not involved in calculating the Etotal from the E of the half reactions. That is it it always Etotal =E1+E2 regarless if you have to mutliply the reactions with a number in order to get the balanced equation.

2. The proper way to write the E for the half reactions would be E0 Mn+2/Mn for the reduction half reaction and
E0 Mn/Mn+2 for the oxidation half reaction. Then
E0 Mn+2/Mn = -E0 Mn/Mn+2
As you said at the anode you have oxidation. But the value of Eanode is by convention the E0 Mn+2/Mn. That's why you get
Ecell =Ecathode-Eanode
(Ecell = E0 Fe+2/Fe + E0 Mn/Mn+2 =E0 Fe+2/Fe - E0Mn+2/Mn =E0 cathode -E0 anode)

3. Having in mind note#2, you could do
Ecell =Ecathode-Eanode and use the values you have for the reduction haf reactions:
Ecell =-1.18-(-0.036) =-1.18+ 0.036 = -1.144 => the reverse is spontaneous => Fe at the cathode => reduction of Fe+3 and oxidation of Mn