Given the following half-reactions, Ag⁺(aq) + e⁻ → Ag(s) E° = +0.799 V, AgCl(s) + e⁻ → Ag(s) + Cl⁻(aq) E° = +0.222 V, calculate the molar solubility of AgCl.
To calculate the molar solubility of AgCl, we need to use the Nernst equation and the information provided to determine the equilibrium constant for the dissolution reaction of AgCl.
The standard electrode potential for the reduction half-reaction involving Ag⁺ is E° = +0.799 V. Considering the balanced equation: Ag⁺(aq) + e⁻ → Ag(s), this half-reaction represents the reduction of Ag⁺ to Ag.
The standard electrode potential for the reduction half-reaction involving AgCl is E° = +0.222 V. Considering the balanced equation: AgCl(s) + e⁻ → Ag(s) + Cl⁻(aq), this half-reaction represents the reduction of AgCl to Ag.
To find the equilibrium constant (Ksp) for the dissolution of AgCl, we can relate the reduction potentials of the two half-reactions using the Nernst equation:
Ecell = E°cell - (0.0592 V / n) * log(Q)
Where:
Ecell is the cell potential,
E°cell is the standard cell potential,
n is the number of electrons transferred in the balanced equation,
Q is the reaction quotient.
In this case, the cell potential (Ecell) is the difference between the reduction potentials of the half-reactions: Ecell = E°Ag⁺ - E°AgCl.
Since Ag⁺ is being reduced to Ag and AgCl is also being reduced to Ag, the number of electrons transferred (n) for both reactions is 1.
Using the Nernst equation, we can set up the expression:
Ecell = +0.799 V - +0.222 V = 0.577 V
Now, we can calculate the value of Q using the formula:
Q = [Ag⁺] / [AgCl]
Since Ag⁺ and Cl⁻ ions come from the dissociation of AgCl, the concentrations of Ag⁺ and Cl⁻ are equal, so [Ag⁺] = [Cl⁻].
Therefore, Q = [Ag⁺]² / [AgCl].
Making use of the relationship between the ion concentration and solubility, we can express [Ag⁺] and [AgCl] in terms of the molar solubility (S):
[Ag⁺] = S and [AgCl] = S.
Now, substituting these values into the equation for Q:
Q = S² / S = S
Since Q = S, we have:
Ecell = 0.577 V = +0.0592 V / 1 * log(S)
To calculate the molar solubility (S), rearrange the equation:
S = 10^(Ecell / (0.0592 V))
S = 10^(0.577 / (0.0592 V))
Evaluating this expression with the given values, we find:
S ≈ 1.12 mol/L
Therefore, the molar solubility of AgCl is approximately 1.12 mol/L.