Reduction and oxidation must occur together. The electrons from the oxidized species are transferred to the reduced species. Chemists often break these two processes apart and write what are referred to as "half reactions." Consider these two half reactions: Zn2+ (aq) + 2 e- → Zn (s) Cu2+ (aq) + 2 e- à Cu (s) a. Are these oxidation or reduction half reactions? b. Calculate ΔG° for each of these reactions. Note that ΔG°f for the electron is 0 kJ/mol. c. Based on your answer to (b), does Zn2+ or Cu2+ more strongly favor reduction?

When a specie tends to lose electrons in a chemical reaction then it means oxidation has taken place. When a specie tends to gain electrons then it means reduction has taken place.

So, for the given reactions the value of [tex]E_{o}[/tex] are as follows.

[tex]Zn^{2+} + 2e^{-} \rightarow Zn[/tex], [tex]E_{o} = -0.76 V[/tex]

[tex]Cu^{2+} + 2e^{-} \rightarow Cu[/tex], [tex]E_{o} = 0.34 V[/tex]

[tex]\Delta G^{o} = -nFE^{o}_{cell}[/tex]

(a) Therefore, according to the given reactions the process is reduction.

(b) Here, n = 2

So, value of [tex]Zn^{2+}[/tex] is as follows.

[tex]\Delta G^{o}_{Zn^{2+}} = -2 \times (-0.76) \times 96500[/tex]

= +146680 V

[tex]\Delta G^{o}_{Cu^{2+}} = -2 \times (-0.34) \times 96500[/tex]

= -65620 V

(c) As per the calculated values, [tex]Zn^{2+}[/tex] is the best reducing agent and [tex]Cu^{2+}[/tex] is the best oxidizing agent.

Also, more positive is the [tex]E_{o}[/tex] value more good it acts as an oxidizing agent. Hence, it is able to favor reduction.

Therefore, [tex]Cu^{2+}[/tex] will more strongly favor reduction.