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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:

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Why do I have to multiply by 2 - what is wrong with my method?

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    $\begingroup$ Well, because oxygen is O2. $\endgroup$ Commented Feb 27, 2021 at 11:34
  • $\begingroup$ @IvanNeretin - I don't quite follow why that is relevant. However, that has made me think: is it because...the half equation for the electrolysis of water is 2H2O --> O2 + 4H+ + 4e- and the half equation for the reduction of lead is Pb2+ + 2e- --> Pb. The ratio of electrons is 2:1...is this why? $\endgroup$
    – vgupt
    Commented Feb 27, 2021 at 11:48
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    $\begingroup$ Well, yes, and now look how much Pb you can buy with those 4e-. $\endgroup$ Commented Feb 27, 2021 at 11:50
  • $\begingroup$ And you really don't need Pv=nRT to do this since the number of moles of oxygen can be calculated directly from the volume of the oxygen (since others have already done the calculation and the volume to number of moles is the same for most gases at STP). $\endgroup$
    – matt_black
    Commented Feb 27, 2021 at 14:01

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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 : $$\ce{2 H2O -> 4H+ + O2 + 4 e-}$$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 $\pu{8.9E-3} ·\pu{207.2 g}$ = $\pu{1.84 g}$

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