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I'm not sure about the specifics of 6M.2 since I did not attempt that problem, but in general you cannot just add standard potentials of reactions because you are usually given the standard reduction potential, which is the potential that stems from a reduction reaction. However, in a redox reaction, one half reaction is oxidation while the other is reduction. Hence, you have to be careful to reverse the sign of the half reaction that involves oxidation before adding them together instead of simply adding the potentials of the half cells together. Also, in 6M.1 and 6M.2, it seems that we are given the overall cell potential and the potential of one half cell, so you need to solve for the potential of the other half cell. Solving the equation cell potential = potential of reduction rxn - potential of oxidation rxn might not necessarily lead to addition. Overall, just be aware of the signs when approaching these problems. Hope this helps!
There are multiple ways to find E. One way is by balancing and adding together the half reactions and in that case you would need to flip the sign of one E˚. However, if you are given E˚ for each half reaction E˚ = E˚ cathode - E˚anode and in that way you simply subtract the standard cell potentials.
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