How can I determine the overall electrochemical reaction that proceeds spontaneously?
Cd2+ | Cd, Co2+ | Co
1. Cd+Co2+ −�¨Cd2++Co
2. Co+Cd2+ −�¨Co2++Cd
(1. and 2. are answer choices)
To determine the overall electrochemical reaction that proceeds spontaneously, you need to consider the standard reduction potentials of the two half-reactions and the Nernst equation.
1. Start by writing the half-reactions for the given electrode setups:
Cd2+ + 2e- -> Cd (reduction half-reaction)
Co2+ + 2e- -> Co (reduction half-reaction)
2. Next, compare the standard reduction potentials (E°) for each half-reaction. The half-reaction with a greater positive E° value has a greater tendency to occur as reduction, while the half-reaction with a smaller positive E° value has a greater tendency to occur as oxidation.
3. The greater tendency for reduction corresponds to the more positive reduction potential. In this case, from the given reduction potentials:
E°(Cd2+ + 2e- -> Cd) = -0.40 V
E°(Co2+ + 2e- -> Co) = -0.28 V
Since -0.28 V is a greater positive value compared to -0.40 V, the half-reaction with Co2+ being reduced to Co has a greater tendency to occur as reduction. Therefore, the overall electrochemical reaction that proceeds spontaneously is option 2:
Co + Cd2+ -> Co2+ + Cd
Look up reduction potentials for
Cd^2+ + 2e==> Cd and
Co2^+ + 2e ==> Co.
Look at which one of the two will give a positive voltage. That is the one that will proceed spontaneously.