CBSE Class 12 · 2003 · Delhi Compartment · Redox Reactions and ElectrochemistryShort Answer

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a) Calculate the electrode potential of a silver electrode dipped in a 0.1 M solution of silver nitrate at 298 K, assuming silver nitrate to be completely dissociated. The standard electrode potential of Ag

^{+}

/Ag is 0.80 V at 298 K.
b) At what concentration of silver ions will this electrode have a potential of zero volt?

Answer: (a) E ≈ 0.741 V (b) [Ag⁺] ≈ 3 × 10⁻¹⁴ M

Solution

(a) Nernst equation (

n = 1

E = E^{\circ} + 0.059 lo g [A g^{+}] = 0.80 + 0.059 lo g (0.1)

= 0.80 - 0.059

E = 0.741 V

(b) Setting

E = 0

0 = 0.80 + 0.059 lo g [A g^{+}]

lo g [A g^{+}] = - \frac{0.80}{0.059} = - 13.56

[A g^{+}] \approx 2.8 \times 1 0^{- 14} M

CBSE Class 12 Board · 2003 · Delhi Compartment · Subjective

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