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Earlier discussion
4 older comments kept for context.
Ojo Sarah
Ojo Sarah
Aderogba Qudus
ΔG = nFe
n = 2
e = 0.042 – 0.012 = 0.03
ΔG = 2×96500×0.03 = 5,790J = 5.79KJ
John Oyekanmi
\( E^{o}{cell} = E^{o}{cathode} -E^{o}{anode} \)
\( E^{o}{cell} = 0.042V -0.012V = 0.03V \)
n(number of electrons) = 2, F (Faraday) = 96500Columbs
∴ using \( \triangle G = -nFE^{o}_{cell} \)
\( \triangle G \) = \( -1 \times 2 \times 96500 \times 0.03 \)
\( \triangle G \) = 5790J = 5.79KJ.
∴ The Correct Answer is B.
John Oyekanmi
\(E^{o}{cell} = E^{o}{cathode} -E^{o}{anode}\)
\(E^{o}{cell}\) = 0.042V -0.012V = 0.03V
n(number of electrons) = 2, F (Faraday) = 96500Columbs
\( \therefore \) using
\( \triangle G = -nFE^{o}_{cell} \)
\( \triangle G = -1 \times2\times96500\times0.03 \)
\( \triangle G \) = 5790J = 5.79KJ
\( \therefore \) The Correct Answer is B.