## Units G=-nFE

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

805312064
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### Units G=-nFE

I'm not sure if this is common sense, but if the units for F are coulombs and the units for E are volts, how do we get joules for G?

Jessica Li 4F
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Joined: Fri Aug 09, 2019 12:16 am

### Re: Units G=-nFE

The units cancel out because volts, coulombs, and joules are all interrelated -- one joule is equal to one volt times one coulomb.

Rodrigo2J
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Joined: Sat Jul 20, 2019 12:16 am

### Re: Units G=-nFE

The units of Volts are actually equivalent to Joules/Coulomb. So, if you work out the math, the units should all cancel and you will get Joules.

Zaynab Hashm 2I
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### Re: Units G=-nFE

1 Volt = 1 J/C so the units cancel out when you multiply;

substituting the units in for -nFE, you get
-(mol)x(C/mol)x(V=J/C), and you’re left with J for G

Nathan Nakaguchi 1G
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### Re: Units G=-nFE

Volts = Joules/Coulomb so in deltaG=-nFE the Coulombs cancel out from the multiplication between E (J/C) and F (C/mol), and n has no units so the final units will be J/mol.

Betania Hernandez 2E
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Joined: Fri Aug 02, 2019 12:15 am

### Re: Units G=-nFE

F is also Coulombs/mol and Volts is Joules/Coulomb. When multiplied together, the units cancel out leaving Joules for deltaG
$-nFE^{\circ}cell=-n(\frac{C}{mol})(\frac{J}{C})$

Diana Chavez-Carrillo 2L
Posts: 122
Joined: Fri Sep 28, 2018 12:18 am

### Re: Units G=-nFE

Zaynab Hashm 2I wrote:1 Volt = 1 J/C so the units cancel out when you multiply;

substituting the units in for -nFE, you get
-(mol)x(C/mol)x(V=J/C), and you’re left with J for G

Does n have units? I have seen some TAs and UAs write mole of e- when plugging in the numbers into n in that equation. But I have also heard that n does not have units and that is why we end up with J/mol as our final units for delta G. Could someone clarify this for me, please?