## 14.11d

Kyra LeRoy 1E
Posts: 52
Joined: Sat Jul 22, 2017 3:00 am

### 14.11d

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)

Aijun Zhang 1D
Posts: 53
Joined: Tue Oct 10, 2017 7:13 am

### Re: 14.11d

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