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CBSE Class 12 · 2004 · Foreign · Redox Reactions and ElectrochemistryShort Answer

Calculate the equilibrium constant for the reaction:

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

(Given:

E_{Z n^{2 +} /Zn}^{\circ} = - 0.763

V and

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

V)

Answer: K ≈ 2.5 × 10³⁷

Solution

E_{ce l l}^{\circ} = 0.34 - (- 0.763) = 1.103 V, n = 2

At 298 K:

lo g K = \frac{n E _{ce l l}^{\circ}}{0.059} = \frac{2 \times 1.103}{0.059} = 37.39

K \approx 2.5 \times 1 0^{37}

CBSE Class 12 Board · 2004 · Foreign · Subjective

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