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  1. I have watch a video about lead acid battery and I am very confused about the way the half reactions are labelled. I have learned that oxydation is the loss of electron (so it's the half reaction with the electrons on the right side) and reduction is the gain of electron (so electrons should be on the left side of the half reaction). This question labels the half reactions the way I would have labelled them, but in the video, the reaction are labelled the other way around. Why?

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  1. I am also confused about the the way the electrode are labelled. I learned that the anode is the electrode where the oxydation occurs (which means the electrode where the electrons leave or where the conventional current enters). This is consistent with this image from wikipedia. But this is not what the video shows, why?

enter image description here

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  • $\begingroup$ Anode is the element where oxidation occurs. So Pb is the anode and it is oxidized. Of course PbO2 is the cathode, and it is reduced. As a consequence, the 2nd drawing is wrong. It shows Pb as the cathode and PbO2 as the anode. PbO2 cannot be oxidized. it cannot be an anode, whatever the nature of the currant (the movement of the electrons is contrary to the sense of the electric currant) $\endgroup$
    – Maurice
    Oct 22 '20 at 19:13
  • $\begingroup$ Thanks for this @Maurice. And what about the first printscreen. $\endgroup$
    – MagTun
    Oct 22 '20 at 19:46
  • $\begingroup$ The first print screen is correct. $\endgroup$
    – Maurice
    Oct 22 '20 at 21:11
  • $\begingroup$ @Maurice, thanks for your help. I have tried again to understand but I am sorry, I don't see it. You wrote that Pb is oxidized and that the first printscreen is right but it indicates that the half reaction with Pb is a reduction (and that PbO2 is oxidized). Why? Thanks again! $\endgroup$
    – MagTun
    Oct 26 '20 at 20:43
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    $\begingroup$ I confess you are right. The first half-equation goes from PbO2 to PbSO4 consuming 2 electrons. So it is a reduction, as Pb goes fro +4 to +2, using 2 electrons. The half-equation is correctly written, but it is a reduction, and not an oxidation. So the name (or title) of this equation is wrong. The second half-equation goes from Pb to PbSO4, releasing 2 electrons; Pb goes from 0 to +2. So it is an oxidation, and not a reduction. The chemical equation is correctly written, but its name (or title) is wrong. So the names (or titles) of the first drawing are wrong. $\endgroup$
    – Maurice
    Oct 26 '20 at 21:15
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Let's speak about the first drawing. The first half-equation goes from $\ce{PbO2}$ to $\ce{PbSO4}$ consuming $2$ electrons. So it is a reduction, as Pb goes from $+4$ to $+2$, using $2$ electrons. The half-equation is correctly written, but it is a reduction, and not an oxidation. The second half-equation goes from $\ce{Pb}$ to $\ce{PbSO4}$, releasing $2$ electrons; $\ce{Pb}$ goes from $0$ to $+2$. So it is an oxidation, and not a reduction. The chemical equation is correctly written, but its name (or title) is wrong. So the names of the first drawing are wrong.

Let's go to the second picture. Anode is the element where oxidation occurs. So Pb is the anode and it is oxidized. Of course $\ce{PbO2}$ is the cathode, and it is reduced. As a consequence, the 2nd drawing is wrong. It shows $\ce{P}$ as the cathode and $\ce{PbO2}$ as the anode. $\ce{PbO2}$ cannot be oxidized. It cannot be an anode, whatever the nature of the currant : the movement of the electrons is contrary to the sense of the electric currant.

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