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If three faradays of electricity is passed through the solutions of AgNO₃ , CuSO₄ and AlCl₃ the mole ratio of cations deposited at the cathodes will be

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a

6 : 3 : 2

b

1 : 1 : 1

c

1 : 2 : 3

d

3 : 2 : 1

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detailed solution

Correct option is A

${AgNO}_{3} \to {Ag}^{+} + {NO}_{3}^{-}; {Ag}^{+} + e^{-} \to Ag$
1 F of electricity deposits 1 mole Ag
$∴$ 3 F gives 3 moles
${CuSO}_{4} \to {Cu}^{+ 2} + {SO}_{4}^{- 2}; {Cu}^{+ 2} + 2 e^{-} \to Cu$
2 F of electricity deposits 1 mole of Cu
$∴$ 3 F gives 1.5 moles
${AuCl}_{3} \to {Au}^{+ 3} + 3 {Cl}^{-}; {Au}^{+ 3} + 3 e^{-} \to Au$
3 F electricity deposits 1 mole of Au
$∴$ ratio of No. of moles of Ag, Cu & Au deposited = 3 : 1.5 : 1 (or) 6 : 3 : 2

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