\(\begin{array}{ll}{\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}} & {E^{\circ}=+0.3 \mathrm{V}} \\ {\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}} & {E^{\circ}=-0.8 \mathrm{V}} \\ {\mathrm{Mn}^{2+}+2 e^{-} \rightarrow \mathrm{Mn}} & {E^{\circ}=-1.2 \mathrm{V}}\end{array}\) Based on the reduction potentials given above, which of the following reactions will be favored? (A) \(\mathrm{Mn}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Mn}+\mathrm{Cu}^{2+}\) (B) \(\mathrm{Mn}^{2+}+\mathrm{Zn} \rightarrow \mathrm{Mn}+\mathrm{Zn}^{2+}\) (C) \(\mathrm{Zn}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Zn}+\mathrm{Cu}^{2+}\) (D) \(\mathrm{Zn}^{2+}+\mathrm{Mn} \rightarrow \mathrm{Zn}+\mathrm{Mn}^{2+}\)

Step by step solution

01

Understand the Reactions

Each of the options given is a redox reaction. The metal on the left side of the equation will be undergoing oxidation (losing electrons and typically forming a cation) while the metal cation on the right will be reduced (gaining electrons to form a neutral atom). For example, in reaction (A), manganese is being oxidized, and copper is being reduced.

02

Analyze Each Option

To identify the favored reaction, the standard cell potential (Ecell) should be determined for each option. For option (A), that would be \(E_{cell}^{A} = E^{o}_{Cu^{2+}/Cu} - E^{o}_{Mn^{2+}/Mn} = +(0.34V) - (-1.18V) = +1.52V\). Do the same calculation for each of the other options.

03

Identify the Reaction with Largest Ecell

The reaction with the highest value of Ecell will be the one that is most favored. Compare all calculated Ecell values and identify the option with the largest Ecell.

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