Using PV = nRT to solve an electrolysis problem Question: Two electrolytic cells are connected in series (the same current passes through each cell). One cell for the electrolysis of water produces $\pu{100 cm^3}$
of oxygen, measured at $\pu{273 K}$ and
$\pu{1.01
×
10^5
 Pa}$
. The second cell contains molten lead(II) bromide,
$\ce{PbBr2}$.
Determine the mass, in $\pu{g}$, of lead produced.
My working: $$PV=nRT$$
$$n={PV\over RT}$$
$$n\pu{={(1.01 \times 10^5)(100 \times 10^{-6})\over (8.31 \times 273)}} $$
$$\pu{= 4.45 \times 10^{-3} moles}$$
$$\pu{mass = 207.2 \times ANS = 0.922 g}$$
The given solution is:

Why do I have to multiply by 2 - what is wrong with my method?
Both methods yield $\pu{4.45E-3}$ mol $\ce{O2}$. But to get $\pu{1 mol}$ $\ce{O2}$, you need $\pu{4 mol}$ electrons, as the half-reaction at the anode is : $$\ce{2 H2O -&gt; 4H+ + O2 + 4 e-}$$So to get $\pu{4.45E-3 mol}$ $\ce{O2}$, the needed amount of electrons is $4$ times the amount of $\ce{O2}$, or $4$· $\pu{4.45E-3}$ = $\pu{1.78E-2}$ electrons. Now $2$ electrons are needed to produce $\ce{1 Pb}$. So $\pu{1.78E-2}$ mol electrons will produce half of it, or $\pu{8.9E-3}$ mole $\ce{Pb}$. And this amount weighs $\pu{8.9E-3} ·\pu{207.2 g}$ = $\pu{1.84 g}$