HW prob 13.11d:Writing the half reactions for galvanic cells

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Mia-Kyla Coronel 2N
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HW prob 13.11d:Writing the half reactions for galvanic cells

Postby Mia-Kyla Coronel 2N » Sat Feb 07, 2015 1:03 am

13.11(d)

Pt(s) | O2(g) | H+ (aq) || OH-(aq) | O2 (g) | Pt (s)

In the solutions manual the end answer is
4H20(l)-->4H+(aq) + 4OH- (aq)

what happened to the O2?

Thank you in advance!

Chem_Mod
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Re: HW prob 13.11d:Writing the half reactions for galvanic c

Postby Chem_Mod » Sat Feb 07, 2015 8:31 am

To solve this problem, balance the following half-reactions from the cell diagram:

O2 —> H+
O2 —> OH-

If you balance the first reaction as a standard reduction reaction in acidic solution, you get:
O2 + 4H+ + 4e- ---> 2H2O.

If you balance the second reaction as a standard reduction reaction in basic solution you get:
O2 + 2H2O + 4e- ---> 4 OH-
In a cell diagram, the anode is always on the left of the double lines and the cathode is on the right. This means the species on the left are involved in the oxidation half-reaction and the species on the right are involved in the reduction half-reaction.

Therefore:
Anode (oxidation): 2H2O ---> O2 + 4H+ + 4e-
Cathode (reduction): O2+ 2H2O + 4e- ---> 4 OH-

Adding the above two half reactions gives the balanced redox reaction: 4H2O(l) ---> 4H+(aq) + 4 OH-(aq)

Joyce 3H
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Re: HW prob 13.11d:Writing the half reactions for galvanic c

Postby Joyce 3H » Sat Feb 07, 2015 5:29 pm

One question on the answer: should we eliminate the coefficient 4 of all the reactants and products of the balanced reaction? Or it doesn't matter? Thank you!

Chem_Mod
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Re: HW prob 13.11d:Writing the half reactions for galvanic c

Postby Chem_Mod » Sat Feb 07, 2015 5:45 pm

Yes, you can eliminate the coefficients and write the balanced equation as: H20(l) --> H+(aq) + OH-(aq)

Shelbyhamilton 3H
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Re: HW prob 13.11d:Writing the half reactions for galvanic c

Postby Shelbyhamilton 3H » Sun Feb 08, 2015 10:00 am

In the standard reduction reaction in the acidic solution, how did you get the H+ onto the reactants side with the O2???

Niharika Reddy 1D
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Re: HW prob 13.11d:Writing the half reactions for galvanic c

Postby Niharika Reddy 1D » Sun Feb 08, 2015 2:48 pm

We initially have O2 →H+

We first balance the 2 oxygens by adding 2H2O:
O2 →H+ + 2H2O

Now, balance the hydrogens by adding H+. There are 5 hydrogens on the right and none on the left, so add 5H+ to the left.
5H+ + O2 →H+ +2H2O

Since H+ appears on both sides, we can simplify by canceling the one on the right out and lowering the number on the left by 1:
4H+ + O2 →2H2O

Add 4 electrons to the left to balance the charges:
4e- + 4H+ + O2 →2H2O

Shelbyhamilton 3H
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Re: HW prob 13.11d:Writing the half reactions for galvanic c

Postby Shelbyhamilton 3H » Sun Feb 08, 2015 3:01 pm

Thank you that is so helpful :) I forgot about that last step!

Victoria Ford 4D
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Re: HW prob 13.11d:Writing the half reactions for galvanic c

Postby Victoria Ford 4D » Sun Feb 07, 2016 1:19 pm

How do you know whether to balance the equation in acidic or basic solution?


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