6.L.3 (d)

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Kevin Xu 4F
Posts: 50
Joined: Fri Aug 09, 2019 12:16 am

6.L.3 (d)

Postby Kevin Xu 4F » Wed Feb 26, 2020 2:48 am

The question asks for the balanced oxidation and reduction half reactions for the hydrolysis of water, reaction given as Pt(s)|O2(g)|H+(aq) ||OH-(aq)|O2(g)|Pt(s). I thought that the compounds listed to the left of the salt bridge, denoted || are the reactants and products of the anode reaction (oxidation) and the compounds listed to the right of the salt bridge are the reactants and products of the cathode reaction (reduction). However, the answer key writes that the
anode reaction is 2H2O --> O2 + 4H+ + 4 e-
cathode reaction is O2 + 2H2O + 4e- -> 4OH-

How come for the anode reaction, both the O2 and the H+ are on the products side? I thought one had to be on the reactants and the other on the products? And if they can be on the same side, how can I tell that from just the cell diagram? Thanks in advance!

Justin Sarquiz 2F
Posts: 106
Joined: Fri Aug 30, 2019 12:15 am

Re: 6.L.3 (d)

Postby Justin Sarquiz 2F » Wed Feb 26, 2020 8:18 am

I had this question too, and my TA said that if you put H+ on one side and 02 on the other, then you cannot balance the equations for the molecules. You can balance the charges for the half reaction, but you cannot have equal number of hydrogen and oxygen molecules on both sides.

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