Question

In: Chemistry

calculate delta g for the reaction between Cr (S) and Cu 2+ (aq)

calculate delta g for the reaction between Cr (S) and Cu 2+ (aq)

Solutions

Expert Solution

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

--------------------------------------------------------------------------------

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)

   


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