CBSE Class 12 · 2009 · All India · Redox Reactions and ElectrochemistryShort Answer

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A copper–silver cell is set up. The copper ion concentration in it is 0.10 M. The concentration of silver ion is not known. The cell potential measured is 0.422 V. Determine the concentration of silver ion in the cell.
(Given:

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

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

Answer: [Ag⁺] ≈ 0.072 M

Solution

Cell: Cu | Cu

^{2 +}

(0.10 M) || Ag

^{+}

(

x

) | Ag,

n = 2

E_{ce l l}^{\circ} = 0.80 - 0.34 = 0.46 V

Nernst equation:

0.422 = 0.46 - \frac{0.059}{2} lo g \frac{0.10}{x ^{2}}

lo g \frac{0.10}{x ^{2}} = \frac{0.46 - 0.422}{0.0295} = 1.288

\frac{0.10}{x ^{2}} = 19.4 \Rightarrow x^{2} = 5.15 \times 1 0^{- 3}

[A g^{+}] = 0.072 M

CBSE Class 12 Board · 2009 · All India · Subjective

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