Given : (i) Zn+4 NH₃ → Zn (NH₃)₄ ²⁺+2 e, E⁰=1.03 V (ii) Zn → Zn²⁺+2 e, E°=0.763 V the formation of the complex Zn (NH₃)₄ ²⁺ is approximately

(c) 1 × 10⁹. Explanation: log β = 2(1.03 − 0.763)/0.0591 ≈ 9: β ≈ 10⁹ (c).

Q91 · CSIR-NET Chemistry, December 2016

Paper: CSIR-NET December 2016 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Medium

Given : (i) $\mathrm{Zn}+4 \mathrm{NH}_{3} \rightarrow \mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}+2 \mathrm{e}, \mathrm{E}^{0}=1.03 \mathrm{V}$ (ii) $\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}, \mathrm{E}^{\circ}=0.763 \mathrm{V}$ the formation of the complex $\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}$ is approximately

(a)$1 \times 10^{5}$

(b)$1 \times 10^{7}$

(c)$1 \times 10^{9}$

(d)$1 \times 10^{12}$

Answer

Answer: C ✓ checked by 4AB · confidence high

Explanation

log β = 2(1.03 − 0.763)/0.0591 ≈ 9: β ≈ 10⁹ (c).

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