Question

The (unbalanced) reduction potentials involved in this reaction are Cu2+ (aq) + 2e- -> Cu(s) E0...

The (unbalanced) reduction potentials involved in this reaction are

Cu2+ (aq) + 2e- -> Cu(s) E0 = 0.34V

NO3- (aq) + 3e- -> NO(g) E0= 0.96V

1)Write out the INDIVIDUAL balanced half reactions, identify which half-reaction is taking place at the anode and cathode and calculate the electrochemical potential of the reaction.

2) Suppose the Ph of the solution was increased so that [H+] = 0.1M, but the concentrations of all other species in solution were unchanged, Would the free energy become more negative, less negative or unchanged?

PLEASE include everything and make sure it is the complete answer. Thank you so much! I will thumbs up if CORRECT!

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