Determine Eo for the following reaction

Question

Determine E^o for the following reaction, using the given standard reduction potentials:

Hg(l) + Mn²⁺(aq) → Hg²⁺(aq) + Mn(s)

E^o for Mn²⁺(aq) = -1.18 V

E^o for Hg²⁺(aq) = +0.85 V

I know its (-1.18) - (0.85) = -2.03

I just dont know why I subtracted and why it's set up that way..

please show steps and explain so I can follow along.

Answer 1

First, find out which is the reduction reaction and which is the oxidation reaction.

Since Eº for Hg²⁺ is greater than that for Mn²⁺, it will be reduced and Mn²⁺ will be oxidized

Recall that reduction takes place at the cathode and oxidation at the anode

Also recall that when using reduction potentials, the Eº_cell = E_cathode - E_anode

Now, just plug in the values

Eº_cell = 0.85 - (-1.18) = 0.85 + 1.18 = 2.03 V