Which among the following equations represents the reduction reaction taking place in lead accumulator at positive electrode, while it is being used as a source of electrical energy?

$P b S O_{4} (s) + 2 e^{-} \to P b (s) + S O_{4}^{2 -} (a q)$

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P b O_{2} (s) + S O_{4}^{2 -} (a q) + 4 H^{+} (a q) + 2 e^{-} \to P b S O_{4} (s) + 2 H_{2} O (l)

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$P b (s) + S O_{4}^{2 -} (a q) \to P b S O_{4} (s) + 2 e^{-}$

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$P b S O_{4} (s) + 2 H_{2} O (l) \to P b O_{2} (s) + S O_{4}^{2 -} + 4 H^{+} + 2 e^{-}$

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Solution

The correct option is B $P b O_{2} (s) + S O_{4}^{2 -} (a q) + 4 H^{+} (a q) + 2 e^{-} \to P b S O_{4} (s) + 2 H_{2} O (l)$
In lead storage battery, lead grids filled with spongy lead will act as anode (negative plate) and lead grids filled with $P b O_{2}$ will act as cathode (positive plate). 38% solution of sulphuric acid will act as the electrolyte for the cell.

When the battery discharge,
At anode: $P b (s) + S O_{4}^{2 -} (a q) \to P b S O_{4} (s) + 2 e^{-}$
Most of the lead sulphate precipitates formed gets sticks to the lead rod. Thus, it is not consumed during discharging.

At cathode: $P b O_{2} (s) + S O_{4}^{2 -} (a q) + 4 H^{+} (a q) + 2 e^{-} \to P b S O_{4} (s) + 2 H_{2} O (l)$
Here also formed lead sulphate precipitate gets sticks
to cathode rod.

Overall reaction during discharge:
$P b (s) + P b O_{2} (s) + 2 H_{2} S O_{4} (a q) \to 2 P b S O_{4} (s) + 2 H_{2} O (l)$
$E_{c e l l} = 2.05 V$

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Similar questions

(a) Here is an electrode reaction :

$C u \to C u^{2 +} + 2 e^{-}$ . At which electrode (anode or cathode) would such a reaction take place? Is this an example of oxidation or reduction ?

(b) A solution contains magnesium ions $(M g^{2 +})$ , iron (II) ions $F e^{2 +}$ and copper ions $C u^{2 +}$ . On passing an electric current through this solution, which ions will be the first to be discharged at the cathode? Write the equation for the cathode reaction.