What is the Cell Potential for Mg/Mg^2+ and Fe/Fe^2+ Electrodes?

[MarcZZ]

Homework Statement

Calculate the cell potential (ECell) of an electrochemical cell with Mg/Mg^2+, and Fe/Fe^2+ electrodes. [Mg^2+] and [Fe^2+] = 1.00 M.

Fe^2+(aq) + 2e^- --> Fe(s) = -0.45V
Mg^2+(aq) + 2e^- --> Mg(s) = -2.37 V

a) 1.90
b) 1.92
c) 1.94
d) 2.82
e) 2.80

Homework Equations

ECell = EReduction + EOxidation

The Attempt at a Solution

I believe the answer is B as I believe because the Mg having a lower value of V will have it's sign changed to form a positive value (oxidation). Is this correct?

 

[Borek]

It is not about changing a sign. It is a about a distance between the half cell potentials - mark them on the number axis and it will be obvious.

Doesn't mean your answer is incorrect.