CBSE Class 11-science Answered

Dear Student,

The equilibrium constant (K) for the reaction,

Sn²⁺ (aq) + Mg(s) → Sn(s) + Mg²⁺(aq)

At the standard state, use the standard electrode potentials for the half-reactions.

The half-reactions and their standard electrode potentials (E∘) are:

Sn²⁺(aq) + 2e−  →  Sn(s)       E_Sn2+/Sn = −0.14V

Mg²⁺(aq) + 2e− → Mg(s)       E_Mg2+/Mg = −2.37 V

In the given reaction, magnesium is oxidised and tin(II) is reduced. Therefore,

Oxidation half-reaction:

Mg(s) → Mg²⁺(aq) + 2e−    E_ox = +2.37V

Reduction half-reaction:

Sn²⁺(aq) + 2e− → Sn(s)       E_red = −0.14 V

The standard cell potential E°cell for the reaction is given by:

E°cell = E°red + E°ox

E°cell = −0.14 V + 2.37 V = 2.23 V

E°cell = −0.14 V + 2.37 V = 2.23 V

The equilibrium constant K is related to the standard cell potential by the Nernst equation at standard conditions (25°C or 298 K): Δ∘ = −E°cell

ΔG∘ = −nFE°cell

ΔG∘ = −RTlnK

Since ΔG∘ is the Gibbs free energy change, equate the RHS of the above two equations to find the value of equilibrium constant K.