Q. 43.8( 47 Votes )

Calculate the sta

Answer :

(i) Known - E0Cr3 + /Cr = - 0.74 V


E0Cd2 + /Cd = - 0.40 V


rG0 = ?


K = ?


The galvanic cell of the given reaction is written as -


Cr(s)|Cr3 + (aq)|| Cd2 + (aq)|Cd(s) Reaction 1


Hence, the standard cell potential is given as,


E0 = E0R - E0L


= - 0.40 - (- 0.74)


E0 = + 0.34 V


To calculate the standard Gibb’s free energy, ∆rG0, we use,


rG0 = - nE0F Equation 1


where nF is the amount of charge passed and E0 is the standard reduction electrode potential.


Substituting n = 6 (no. of e - involved in the reaction 1), F = 96487 C mol-1,


E0 = + 0.34 V in Equation 1, we get,


rG0 = - 6×0.34V×96487 C mol-1


= - 196833.48 CV mol-1


= - 196833.48 J mol-1


rG0 = - 196.83348 kJ mol-1


To find out the equilibrium constant, K, we use the formula,




log K = 34.5177


K = antilog 34.5177


K = 3.294 × 1034


The standard Gibb’s free energy, ∆rG0 is - 196.83348 kJ mol – 1 and equilibrium constant, K is 3.294 × 1034


(ii) Known -


E0Fe3 + /Fe2 + = 0.77V


E0Ag + /Ag = 0.80 V


rG0 = ?


K = ?


The galvanic cell of the given reaction is written as -


Fe2 + (aq)|Fe3 + (aq)|| Ag + (aq)|Ag(s) Reaction 1


Hence, the standard cell potential is given as,


E0 = E0R - E0L


= 0.80 - (0.77)


E0 = + 0.03 V


To calculate the standard Gibb’s free energy, ∆rG0, we use,


rG0 = - nE0F Equation 1


where nF is the amount of charge passed and E0 is the standard reduction electrode potential.


Substituting n = 1 (no. of e - involved in the reaction 1), F = 96487 C mol-1, E0 = + 0.03V in Equation 1, we get,


rG0 = - 1×0.03V×96487 C mol-1


= - 2894.61 CV mol - 1


= - 2894.61 J mol-1


rG0 = - 2.89461 kJ mol-1


To find out the equilibrium constant, K, we use the formula,




log K = 0.5076


K = Antilog 0.5076


K = 3.218


The standard Gibb’s free energy, ∆rG0 is - 2.89461 kJ mol – 1 and equilibrium constant, K is 3.218


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