Why is w(max) = delta G?

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

Viraj B 3A
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Why is w(max) = delta G?

In formulating the equation: $\Delta$G = -nFE, how can we assume that wmax = $\Delta$G under constant temperature and pressure? Thank you for any insight.

Chem_Mod
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Re: Why is w(max) = delta G?

G is actually defined as the maximum non-expansion work under constant T and P. Non-expansion means not related to a volume change, for example, electrical work, which is also what we call "useful work". This can be proven mathematically using differentials, obviously you won't be responsible for this!

$dU = dq + dw_{expansion} + dw_{non expansion}$ 1st Law, split the work into expansion and non-expansion
$dU = dq - PdV + dw$ Expansion work is -PdV, assume from now that dw refers to non-expansion
$dU < TdS - PdV + dw$ Clausius inequality: dS > dq/T so TdS > dq
$dU + PdV - TdS < dw$ Move terms
$dU + (PdV + VdP) - (TdS + SdT) < dw$ We can arbitrarily add VdP and SdT terms. They are zero since T, P constant
$dU + d(PV) - d(TS) < dw$ Product rule for derivatives
$d(U+PV-TS) < dw$
$dG < dw$ because G = H-TS = U+PV-TS

Recall that work done by the system is negative, so if DeltaG < w, then DeltaG is the maximum work that may be done.