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CBSE Class 12 · 2004 · Delhi Compartment · Redox Reactions and ElectrochemistryShort Answer

The emf of the cell

Zn ∣ Z n^{2 +} (0.1 M) ∣∣ C d^{2 +} (M_{1}) ∣ Cd

has been found to be 0.3305 V at 298 K. Calculate the value of M

_{1}

. Given:

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

V,

E_{C d^{2 +} /Cd}^{\circ} = - 0.40

V.

Answer: M₁ = 0.01 M

Solution

E_{ce l l}^{\circ} = - 0.40 - (- 0.76) = 0.36 V, n = 2

Nernst equation:

0.3305 = 0.36 - \frac{0.059}{2} lo g \frac{0.1}{M _{1}}

lo g \frac{0.1}{M _{1}} = \frac{0.36 - 0.3305}{0.0295} = 1

\frac{0.1}{M _{1}} = 10

M_{1} = 0.01 M

CBSE Class 12 Board · 2004 · Delhi Compartment · Subjective

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