13.9 b, the n in Gibbs free energy

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Madison Davis 3F
Posts: 44
Joined: Fri Sep 26, 2014 2:02 pm

13.9 b, the n in Gibbs free energy

Postby Madison Davis 3F » Fri Feb 06, 2015 2:37 pm

Hello,

For the equation
6Fe(3+) +3Cr(3+) +7H20 --> 6Fe(2+) +Cr2O7(2-) + 14H(+)

How do you break this equation up into 1/2 reactions?
Is this how you find out the n in the equation -nFE, by calculating the electrons added? (Please correct me if I'm wrong)

Also, wouldn't the electrons added be different for each half reaction? If so, which one would you use for your n in the equation -nFE?

Thank you for your help!

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

Re: 13.9 b, the n in Gibbs free energy

Postby Niharika Reddy 1D » Fri Feb 06, 2015 3:12 pm

Since the equation given is already balanced (the total charge and number of each element are the same on both sides), either half reaction you look at will give you the same number of moles of electrons transferred (n). To split it up into half reactions, you can assign oxidation states to see where there's an increase in oxidation number (oxidation) or decrease in oxidation number (reduction). Since it's already balanced, it suffices to pick just one of the two. In the reaction, Fe3+ is being reduced since its oxidation number decreases from +3 to +2, and since there are 6 moles of Fe3+ in the given balanced reaction, the reduction half reaction had to be multiplied by 6 in order to obtain the balanced reaction. Multiplying by 6 would mean 6 electrons were gained during reduction, so n=6.
6[Fe3+ + e- --> Fe2+]

Since E°cell is given, all we needed to find was n, as now we can use ΔG°=-nFE° to calculate the standard Gibbs free energy.


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