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Chapter 10: Problem 22

Which of the following reactions is spontaneous? (a) \(3 \mathrm{Zn}^{2+}+2 \mathrm{Al} \rightarrow 3 \mathrm{Zn}+2 \mathrm{Al}^{3+}\) (b) \(3 \mathrm{Zn}+2 \mathrm{Al}^{3+} \rightarrow 3 \mathrm{Zn}^{2+}+2 \mathrm{Al}\)

Short Answer

Expert verified
Neither reaction (a) nor (b) is spontaneous, as both have negative potential differences. ΔE (a) = -2.439 V and ΔE (b) = -0.913 V.

Step by step solution

01

Determine the reduction potentials of each half-reaction

For this, we can refer to the standard reduction potential table. The potentials are as follows: Reduction of Zn²⁺: Zn²⁺ + 2 e⁻ → Zn E° = -0.763 V Oxidation of Al: Al → Al³⁺ + 3 e⁻ E° = -1.676 V (We can simply change the sign, because the table gives reduction potentials and here, Al is being oxidized). Reduction of Al³⁺: Al³⁺ + 3 e⁻ → Al E° = -1.676 V Oxidation of Zn: Zn → Zn²⁺ + 2 e⁻ E° = 0.763 V (We change the sign for the same reason mentioned above).
02

Calculate the potential difference (ΔE) for reaction (a)

ΔE = E° (reduction) + E° (oxidation) ΔE (a) = E° (3 Zn²⁺ + 6 e⁻ → 3 Zn) + E° (2 Al → 2 Al³⁺ + 6 e⁻) ΔE (a) = (-0.763 V) + (-1.676 V) = -2.439 V
03

Calculate the potential difference (ΔE) for reaction (b)

ΔE = E° (reduction) + E° (oxidation) ΔE (b) = E° (2 Al³⁺ + 6 e⁻ → 2 Al) + E° (3 Zn → 3 Zn²⁺ + 6 e⁻) ΔE (b) = (-1.676 V) + (0.763 V) = -0.913 V
04

Identify the spontaneous reaction

The reaction with the highest positive potential difference (ΔE) is the spontaneous one. In this case, both reactions have a negative potential difference, meaning neither of them is spontaneous. So, neither (a) nor (b) is a spontaneous reaction.

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