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CBSE Class 12 · 2011 · All India Compartment · Redox Reactions and ElectrochemistryShort Answer

Calculate the equilibrium constant K for the reaction at 298 K:

Zn (s) + C u^{2 +} (aq) ⇌ Z n^{2 +} (aq) + Cu (s)

(Given:

E_{Z n^{2 +} /Zn}^{\circ} = - 0.76

V,

E_{C u^{2 +} /Cu}^{\circ} = + 0.34

V)

Answer: K ≈ 2 × 10³⁷

Solution

E_{ce l l}^{\circ} = 0.34 - (- 0.76) = 1.10 V, n = 2

lo g K = \frac{n E _{ce l l}^{\circ}}{0.059} = \frac{2 \times 1.10}{0.059} = 37.29

K \approx 1.9 \times 1 0^{37} \approx 2 \times 1 0^{37}

CBSE Class 12 Board · 2011 · All India Compartment · Subjective

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