## Delta G=-nFE

$\Delta G^{\circ} = -nFE_{cell}^{\circ}$

John Huang 1G
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### Delta G=-nFE

A lot of the problems do not have moles given, so when we are using DeltaG=-nFE, what value would we use for n?

Dylan Mai 1D
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### Re: Delta G=-nFE

do you have an example of one of the problems that does this?

Rachel Formaker 1E
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### Re: Delta G=-nFE

Moles in the equation ΔG=-nFE is for moles of electrons, so even if the moles of a substance are not explicitly given, you should be able to figure out the moles of electrons through the redox half-reactions.

Julie Steklof 1A
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### Re: Delta G=-nFE

After writing balanced half reactions, you will be able to see the moles of electrons involved in the redox reaction. This is the number you use for n.

Ivy Lu 1C
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### Re: Delta G=-nFE

The n in the equation is the moles of electrons transferred. To find the electrons transferred, you'll have to write the two half-reactions and balance them.

Jeremiah Samaniego 2C
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### Re: Delta G=-nFE

Ivy Lu 1C wrote:The n in the equation is the moles of electrons transferred. To find the electrons transferred, you'll have to write the two half-reactions and balance them.

A good example of this is homework problem 14.9.a):

Given the overall equation 2 Ce4+ (aq) + 3 I- (aq) --> 2 Ce3+ (aq) + I3- (aq), the half reactions would be:
2 Ce4+ (aq) + 2 e- --> 2 Ce3+ (aq) and 3 I- (aq) --> I3- (aq) + 2 e-.

Since 2 moles of electrons are being transferred in the overall reaction, you would plug in 2 for n in the equation Delta G = -nFE