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
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?