Question

In: Chemistry

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

Expert Solution

anode reaction: oxidation takes place

Cr(s) -------------------------> Cr+2 (aq) + 2e- , E⁰_Cr+2/Cr = - 0.91V

cathode reaction : reduction takes palce

Cu+2(aq) + 2e- -----------------------------> Cu(s) , E⁰_Fe+2/Fe = +0.34V

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

net reaction: Cr(s) +Cu+2(aq) -------------------------> Cr+2 (aq) + Cu(s)

if standard reduction potentiala are given then

E⁰_cell= E⁰_cathode- E⁰_anode

E⁰_cell= E⁰_Cu+2/Cu - E⁰_Cr+2/Cr

= 0.34 - (-0.91)

= 1.25V

Gibbs free energy calculation

detla G⁰ = -nF E_cell⁰ ---------------------------(1)

here n= number of moles of electrons transfered =2

F = Faraday = 96500 columbs

substitute all values in (1)

detla G⁰ = -nF E_cell⁰

= -2 x 96500 x 1.25

=- 241250J

detla G⁰ = - 241.25 kJ ---------------------(answer)

queen_honey_blossom answered 9 months ago

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