6M.5

[vanessas0123]

In the cell diagram Hg|Hg2^2+||NO3-,H+|NO|Pt, the half reactions are like this:

3e- + NO3- -> NO (red potential: 0.96V)
2Hg -> Hg2^2+ + 2e-

However, only the reduction potential of 2e- + Hg2^2+--> 2Hg is given (0.79V). Why don't we flip the sign since its supposed to be an oxidation reaction (0-->1+)? If you flip the sign, you won't get the correct answer (0.96-0.79=0.17V)

[Sebastian Lee 1L]

Remember that you can only use one method to find the E_cell. If you use the equation , you do NOT flip any signs and only use the reduction potentials as given in the chart. In your example, you would find the reduction potential of NO₃^- (0.96V) and subtract the reduction potential given for Hg₂²⁺ (0.79 V). So 0.96-0.79=0.17V as the answer. The other method is to flip the E0 of the anode like you did and then just add that to the other E0 of the cathode. In doing so, your flipped anode potential of -0.79 V is added to the 0.96 V potential cathode which gives you the same answer.

Again, you can't combine the two methods where you flip the sign on the anode and then subtract that number. Only choose one.