How Do You Determine Which Reaction to Flip in a Voltaic Cell?

[a.a]

Homework Statement

A voltaic cell at 25oC consists of Mn/Mn2+ and Cd/Cd2+ half-cells with an initial cell potential of +0.768 V. The [Mn2+] concentration is 0.500 M. Use the Nernst equation to calculate the [Cd2+] concentration in the Cd/Cd2+ half-cell.
Cd+2(aq) + 2e- = Cd(s) . . . . . Eo = -0.40 V
Mn+2(aq) + 2e- = Mn(s) . . . . . Eo = -1.18 V

Homework Equations

Ecell = E(standard) cell - (RT/nF)ln Q

The Attempt at a Solution

I was just wondering how do we know which reaction above, we need to flip and how do we determine which metal is oxidized and reduced in cells like this? Is it that the half-cell woth the more negative potential is more redilly reduced, thus that's the one you flip, to create the cell?
I know how to use the nerst equation and find concentration, but am having trouble with Enot Cell because i don't know which eqn to flip.

 

[Borek]

What and why do you want to flip?

All you need is

E_Mn/Mn²⁺ = E_Cd/Cd²⁺

 

[a.a]

How is EMn/Mn2+ = ECd/Cd2+ ?
EMn/Mn2+ = -1. 18 and ECd/Cd2+= -0.40

The potential given is the E(reduction), if we use the equation E(cell) = E(oxidation) + E(reduction). Then we need to change the sign of one of the potentials to get that reactions E(oxidation).