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what is the equilibrium constant k
Asked by anithaanu629940 | 23 May, 2024, 08:33: AM

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)       ESn2+/Sn = −0.14V

Mg2+(aq) + 2e− → Mg(s)       EMg2+/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−    Eox = +2.37V

Reduction half-reaction:

Sn2+(aq) + 2e− → Sn(s)       Ered = −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.

Answered by | 23 May, 2024, 10:53: AM
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