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Faraday's Second Law

Exactly 0.4...

Question

Exactly $0.4$ faraday electric charge is passed through three electrolytic cells in series, first containing $A g N O_{3}$ , second $C u S O_{4}$ and third $F e C l_{3}$ solution. How many grams of each metal will be deposited assuming only cathodic reaction in each cell?

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Solution

The cathodic reactions in the cells are respectively,
$A g^{+} 1 m o l e 108 g + e^{-} 1 m o l e 1 F \to A g$
$C u^{2 +} 1 m o l e 63.5 g + 2 e^{-} 2 m o l e 2 F \to C u$
and $F e^{3 +} 1 m o l e 56 g + 3 e^{-} 3 m o l e 3 F \to F e$
Hence, $A g$ deposited $= 108 \times 0.4 = 43.2 g$
$C u$ deposited $= \frac{63.5}{3} \times 0.4 = 12.7 g$
and $F e$ deposited $= \frac{56}{3} \times 0.4 = 7.47 g$

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