[opti384]

I have the reaction 2Cr(s) + 3Cu^2+ (aq) → 2Cr^3+ (aq) + 3Cu(s) E^° = 0.43 V. and I have to find the
∆G^°  in KJ mol ^-1. So I came across the equation ∆G^° = -nFE^°. The answer is -6 × 96500 × 0.43  / 1000. At first I though it was -6 × 96500 × 0.43. I think the /1000 comes because I have to find ∆G^°  in KJ mol ^-1. Is ∆G^° originally expressed in J mol ^-1?

[MrTeo]

Yes, because

$$ 1\;V=\frac{1\;J}{1\;C} /$$

F is measured in Coulombs (it's the charge of a mole of electrons) and n is adimensional.

[opti384]

I see.