Half Equation in electrochemical Cells

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Prismere
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I know the cell emf is -0.4--2.37 = 1.93 V

However, Reduction happens at the Cathode (negative electrode): REDCAT.
Fe2+ + 2e- ->Fe is more positive than the Mg2+ half equation, thus that oxidises the Mg.

So why is it that the answer is "Half equation:Mg → Mg2+ + 2 e (1) "
and not Fe2+ + 2e- -> Fe
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charco
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(Original post by Prismere)
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I know the cell emf is -0.4--2.37 = 1.93 V

However, Reduction happens at the Cathode (negative electrode): REDCAT.
Fe2+ + 2e- ->Fe is more positive than the Mg2+ half equation, thus that oxidises the Mg.

So why is it that the answer is "Half equation:Mg → Mg2+ + 2 e (1) "
and not Fe2+ + 2e- -> Fe
You have assumed that the negative electrode is the cathode. This is not the case for electrochemical cells. Oxidation always occurs as the anode and reduction at the cathode.

In an electrochemical cell the electrons are produced at the negative electrode (anode) and flow around the external circuit to the positive electrode (cathode)
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