What are the oxidizing and reducing agents in the reaction? $$\ce{Pb(s) + PbO2(s) + 2H^+ + 2HSO_4^- -> 2PbSO4(s) + 2H2O(l)}$$

I thought that, since the solid lead is going from $0$ to $2+$ it is being oxidized, and in the half reaction it works with the $\ce{HSO4^-}$, so the $\ce{HSO4^-}$ would be the oxidizing agent.

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    $\begingroup$ Sulfur can have different oxidation states, but it is $\ce{SO4^{2-}}$ on both sides of the equation so it isn't being oxidized or reduced. $\endgroup$ – MaxW Mar 15 '16 at 3:43

You are correct, the Pb(s) is oxidized, which makes it the reducing agent. However, there is no change in the oxidation state of the H or the SO4−, so it is not the oxidizing agent. Rather the PbO2(s) is the oxidizing agent, as the Pb goes from +4 in the PbO2(s) to +2 in the PbSO4(s).


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