(i) E^⊖_Cr³⁺/Cr = 0.74 V

E^⊖_Cd²⁺/Cd  = -0.40 V

The galvanic cell of the given reaction is depicted as:

Cr(s)|Cr³⁺(aq)||Cd²⁺(aq)|Cd(s)

Now, the standard cell potential is

E^⊖_Cell = E^⊖_R - E^⊖_L

= -0.40 - (-0.74)

= +0.34v

Δ_rG^⊖ = -nFE^⊖_Cell

In the given equation,

n = 6

F = 96487 C mol^-1

E^⊖_Cell = +0.34 V

Then, Δ_rG^⊖ = - 6 × 96487 C mol^-1 × 0.34 V

= - 196833.48 CV mol^-1

= - 196833.48 J mol^-1

= - 196.83 kJ mol^-1

Again,

Δ_rG^⊖ = -RTlnK

⇒ Δ_rG^⊖ = -2.303 RTlnK

⇒ log K = \(\frac{Δ_rG}{2.303 RT}\)

= \(\frac{-196.83 \times 10^3}{2.303 \times 8.314 \times 298}\)

= 34.496

∴ K = antilog (34.496)

= 3.13 × 10³⁴

(ii) E^⊖_{Fe³⁺/Fe²⁺} = 0.77 V

E^⊖_Ag⁺/Ag = 0.80 V

The galvanic cell of the given reaction is depicted as:

Fe²⁺(aq)|Fe³⁺(aq)||Ag⁺ (aq)|Ag(s)

Now, the standard cell potential is

E^⊖_Cell = E^⊖_R - E^⊖_L

= 0.80 - 0.77

= 0.03 V

Here, n = 1.

Then, Δ_rG^⊖ = -nFE^⊖_Cell

= - 1 × 96487 C mol^-1 × 0.03 V

= - 2894.61 J mol^-1

= - 2.89 kJ mol^-1

Again, Δ_rG^⊖ = -2.303 RTlnK

⇒ log K = \(\frac{Δ_rG}{2.303 RT}\)

= \(\frac{-2894.61}{2.303 \times 8.314 \times 298}\)

= 0.5073

∴ K = antilog (0.5073)

= 3.2 (approximately)