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As an anode loses electrons, the electrolyte will react with the anode producing lead sulfate.

Simultaneously, the cathode, receives electrons from the circuit which came from the anode, and will give up electrons to the electrolyte and produce lead sulfate.

Is this true that both electrodes produce lead sulfate during lead acid battery discharge?

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    $\begingroup$ So what is the chemical reaction? A quick google should help you. BTW it is a reversible reaction. $\endgroup$
    – Solar Mike
    Aug 11 at 7:03
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Negative plate reaction:

$$ \mathrm{Pb}(s) \ + \ \mathrm{HSO_4}(aq)\ {\xrightarrow{\text{discharging}}}\ \mathrm{PbSO_4} (s) \ +\ \mathrm{H^+} (aq)\ +\ 2e^-$$

Positive plate reaction:

$$ \mathrm{PbO_2}(s) \ + \ \mathrm{HSO_4}(aq) \ + \ \mathrm{3H^+} (aq) \ +\ 2e^-\ \xrightarrow{\text{discharging}}\ \mathrm{PbSO_4} (s) \ +\ \mathrm{2H_2O} (l)$$

It's what the chemical reactions say. On the cathode, lead plates become lead sulfate and the anode, lead dioxide plate becomes lead sulfate. Sulfuric acid concentration decreases and water concentration increases (electrolyte becomes less acidic).

Process reverses during charging.

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