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CBSE Class 12 · 2005 · All India Compartment · Redox Reactions and ElectrochemistryShort Answer

Calculate the emf of the following cell:

Mg (s) ∣ M g^{2 +} (0.2 M) ∣∣ A g^{+} (1 \times 1 0^{- 3} M) ∣ Ag (s)

E_{A g^{+} /Ag}^{\circ} = 0.80

V,

E_{M g^{2 +} /Mg}^{\circ} = - 2.37

V

Answer: E_cell ≈ 3.01 V

Solution

Cell reaction: Mg + 2Ag

^{+} \to

Mg

^{2 +}

+ 2Ag,

n = 2

:

E_{ce l l}^{\circ} = 0.80 - (- 2.37) = 3.17 V

Nernst equation:

E_{ce l l} = 3.17 - \frac{0.059}{2} lo g \frac{0.2}{( 1 0 ^{- 3} ) ^{2}}

= 3.17 - 0.0295 lo g (2 \times 1 0^{5}) = 3.17 - 0.0295 \times 5.30

E_{ce l l} = 3.01 V

CBSE Class 12 Board · 2005 · All India Compartment · Subjective

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