Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
(i) 2Cr(s) + 3Cd2+ (aq) → 2Cr3+ (aq) + 3Cd
(ii) Fe2 + (aq) + Ag+ (aq) → Fe3+ (aq) + Ag(s)
Calculate the ΔrG0 and equilibrium constant(K) of the reactions.
(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