In: Chemistry
calculate delta g for the reaction between Cr (S) and Cu 2+ (aq)
anode reaction: oxidation takes place
Cr(s) -------------------------> Cr+2 (aq) + 2e- , E0Cr+2/Cr = - 0.91V
cathode reaction : reduction takes palce
Cu+2(aq) + 2e- -----------------------------> Cu(s) , E0Fe+2/Fe = +0.34V
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net reaction: Cr(s) +Cu+2(aq) -------------------------> Cr+2 (aq) + Cu(s)
if standard reduction potentiala are given then
E0cell= E0cathode- E0anode
E0cell= E0Cu+2/Cu - E0Cr+2/Cr
= 0.34 - (-0.91)
= 1.25V
Gibbs free energy calculation
detla G0 = -nF Ecell0 ---------------------------(1)
here n= number of moles of electrons transfered =2
F = Faraday = 96500 columbs
substitute all values in (1)
detla G0 = -nF Ecell0
= -2 x 96500 x 1.25
=- 241250J
detla G0 = - 241.25 kJ ---------------------(answer)