14.11 d)

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14.11 d)

Postby AshleyAguilar_1F » Mon Feb 13, 2017 6:38 pm

So for the right side of the cell diagram the half-reaction should be O2 (g) + 2H2O (l) + 4e- --> 4OH - (aq) . **originally just O2 (g) --> OH- (aq)** I don't understand why the water molecules are on the reactants side if the first step to balancing the equation is to balance the oxygen molecules (so adding H2O to the products side, then adding 3H+ to the reactants to balance the hydrogen). So basically what is the process of balancing the half-reaction equations in these kinds of problems

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Re: 14.11 d)

Postby Tara_Hekmati_3B » Mon Feb 13, 2017 6:50 pm

O2 (g) --> OH- (aq)
Basically when you look at this half reaction you want to try to balance the oxygens on both sides because there are two on the left and one on the right. You try doing this by doubling the about of OH-. Now you have...
O2 (g) --> 2OH- (aq)
Then you want to balance the hydrogens by using H2O. So you add it to the left side...
O2 (g) + H2O (l) --> 2OH- (aq)
Then you realize that there are 3 oxygens on the left side and only an even number of oxygens are permitted on the right side. So changing the coefficients on either side will not solve our problem. So we need to rebalance the oxygen atoms by giving the left side an even number of oxygen atoms, and changing the right side accordingly because the hydrogens will then be unbalanced...
O2 (g) + 2H2O (l) --> 2OH- (aq)
and then...
O2 (g) + 2H2O (l) --> 4OH- (aq)
Next balance the charges by adding electrons to the left side...
O2 (g) + 2H2O (l) + 4e- --> 4OH - (aq)

Aaaaand done.

Helen Shi 1J
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Re: 14.11 d)

Postby Helen Shi 1J » Wed Feb 21, 2018 11:59 pm

Can someone explain how do part d, especially writing the half redox reaction for the left side? How did we know H2O was the reactant for the left but O2 the reactant for the right?

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