An electric current is passed through an aqueous solution of lead(II) chloride. What are the half-reactions that take place at the cathode and the anode?

C a t h o d e - P b^{2 +} + 2 e^{-} \to P b, A n o d e - C l^{-} \to C l + e^{-}

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A n o d e - P b^{2 +} + 2 e^{-} \to P b, C a t h o d e - C l \to C l^{+} + e^{-}

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c a t h o d e - P b^{2 +} \to P b^{+} 2 e^{-}, a n o d e - C l^{-} \to C l + e^{-}

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A n o d e - P b \to P b^{2 +} 2 e^{-}, C a t h o d e - C l^{-} \to C l + e^{-}

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Solution

The correct option is A $C a t h o d e - P b^{2 +} + 2 e^{-} \to P b, A n o d e - C l^{-} \to C l + e^{-}$
We know the process of oxidation takes place at the anode and that of reduction at the cathode.
The compound $P b C l_{2}$ renders the ions $P b^{2 +}$ and $C l^{-}$ , so $P b^{2 +}$ gets reduced and $C l^{-}$ gets oxidized.
So at anode the reaction that takes place is:
$C l^{-} \to C l + e^{-}$ (oxidation)
$P b^{2 +} + 2 e^{-} \to P b$ (reduction at cathode)

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