6L.3 d)

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Jordan Young 2J
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6L.3 d)

Postby Jordan Young 2J » Wed Feb 26, 2020 4:59 pm

Write the half-reactions and the balanced equation for the cell reaction for each of the following galvanic cells:

The answer is: 2H2O --> 4H++ O2 + 4e- for the anode and O2 + 2H2O + 4e- --> 4OH- for the cathode

How do you know that H+ and O2 are both products for the anode, and that O2 and OH- are on opposite sides for the cathode?

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Re: 6L.3 d)

Postby AArmellini_1I » Wed Feb 26, 2020 6:05 pm

I believe it because you know the anode is an oxidation. And since you only have O2 and H+ for the anode the only reasonable reaction you can make were oxidation occurs would be having 2H2O --> 4H+ + 2OH- + 4e-. And same for reduction in the cathode. Since you only have OH- and O2 the only reasonable reaction can be O2 + 2H2O + 4e- --> 4OH-

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