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14.11d

Posted: Sun Feb 18, 2018 3:50 pm
by Kyra LeRoy 1E
Can someone walk me through writing the half-equations and balanced equation for this problem?

Pt(s)|O2(g)|H+(aq)||OH-(aq)|O2(g)|Pt(s)

Re: 14.11d

Posted: Sun Feb 18, 2018 4:56 pm
by Aijun Zhang 1D
Pts are just inert electrodes.
The oxygen and H+ ions are on the anode side. Oxidation occurs anode side. So oxygen in the water will be oxidized into oxygen.

On the cathode side, reduction occurs. So O2 is reduced to OH-.


Then we need to balance the number of electrons on both sides. Since there are 4e- on both sides, we just need to add them up and subtract the same substance.
Then the balanced equation will be :