Reactions for the lead acid battery are:
$$ \begin{array}{} \text{Oxidation}&\ce {Pb(s) + HSO4^-(l) &-> PbSO4(s) + H+(l) + 2e-}\\ \text{Reduction}&\ce{PbO2 + HSO4^-(l) + 3H+(l) + 2e- &-> PbSO4(s) + 2H2O}\\ \text{Total reaction}&\ce{Pb(s) + PbO2(s) +2HSO4^- +2H+ &-> PbSO4 + 2H2O}\\ \end{array} $$
What will be the cell notation for this battery? My attempt:$$\ce{Pb(s), Pb^2+(s)| HSO4^-(l)| PbO2(s), Pb^2+(s),Pb(s)}$$
The things that I have in mind,
- In both sides the electrode material is $\ce{Pb(s)}$
- Separate every element that is in the same phase with a comma
- Separate every element with different phases with a single bar
- Separate the two half reactions with a single bar (as I find no salt bridge here)
I am having a tough time learning cell notation. Anyways this was a challenging one for me and I tried my best shot. I am really not sure whether everything is right or not. So please identify my mistakes and show me the right way to do it.
Answer
There are a couple of things wrong here. First off, your final reaction is unbalanced. Once you've fixed the balancing, read the other mistakes:
- The ions do not exist in the liquid state! They are solvated/hydrated by the solvent. Since the solvent is water here, we'll say that the ions are in the aqueous (aq) phase instead.
- While there is certainly no salt bridge here, there is still an electrolyte - an aqueous solution of sulphuric acid. Hence, you must mention it in your cell notation, between the anode and the cathode.
- The $\ce{PbSO4}$ formed at the anode is in solid state. Hence, writing it as $\ce{Pb^2+(s)}$ is incorrect, as it is not dissociated into the ions $\ce{Pb^2+}$ and $\ce{SO4^2-}$.
With all these corrections, your final, correct cell representation should be:
$$\small{\ce{Pb(s), HSO4^-(aq) | PbSO4(s), H+ | H2SO4 ($\pu{x~M}$) | PbO2(s), HSO4-(aq), H+(aq) | PbSO4(s)}}$$
Some websites (like KhanAcademy) and texts (NCERT 12), cite the lead-acid battery reaction as this instead:
$$ \begin{array}{} \text{Oxidation}&\ce {Pb(s) + SO4^2-(aq) &-> PbSO4(s) + 2e-}\\ \text{Reduction}&\ce{PbO2 + SO4^2-(aq) + 4H+(aq) + 2e- &-> PbSO4(s) + 2H2O(l)}\\ \text{Total reaction}&\ce{Pb(s) + PbO2(s) +2SO4^2(aq)- + 4H+(aq) &-> 2PbSO4(s) + 2H2O(l)}\\ \end{array} $$
assuming the bisulphite ion to be further ionized. The cell notation in this case would then be:
$$\small{\ce{Pb(s), SO4^2-(aq) | PbSO4(s) | H2SO4 ($\pu{x~M}$) | PbO2(s), SO4^2-(aq), H+(aq) | PbSO4(s)}}$$
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