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### 14.11d

Posted: Sun Feb 18, 2018 3:50 pm
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
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.
$2H_{2}O \rightarrow 4H^{+}+O_{2}+4e^{-}$
On the cathode side, reduction occurs. So O2 is reduced to OH-.
$O_{2}+2H_{2}O+4e^{-}\rightarrow 4OH^{-}$

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 : $H_{2}O\rightarrow H^{+}+OH^{-}$