Answer to Question #114826 in General Chemistry for kirthana

Question

Cr_2 O_7^(2-)+14H^++6e^-⟶Cr^(+++)+7H_2 O,E^∘=1.33V:3×[2I^-…I_2+2e^(-1)].E^0=-0.54V Find out the value of the equilibrium constant and Gibbs free energy change in the reaction given above.

Answer 1

Answer :

K=1.596×1080,ΔG∘=457.41kJmol-1

Solution :E∘cell=1.33+(-0.54)V=0.79V,n=6.