Question

When [Ag⁺] = 1.22 M, the observed cell potential at 298 K for an electrochemical cell with the reaction shown below is 2.527 V. What is the Mg²⁺ concentration in this cell?

2Ag⁺ (aq) + Mg(s) + 2Ag(s) + Mg²⁺ (aq)

[Mg²⁺] = ______ mol/L

Answer 1

The concentration of Mg²⁺ in the cell is approximately 1.01 * 10^-4 M.

Step-by-step explanation:

The Nernst equation can be used to calculate the standard electrode potential for the reaction that occurs at the cathode. The Nernst equation is given by: E = E° - (RT/nF) * ln([Mg²⁺]/[Ag⁺]²), where E is the observed cell potential, E° is the standard electrode potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the reaction, F is Faraday's constant, and [Mg²⁺] and [Ag⁺] are the concentrations of Mg²⁺ and Ag⁺ ions, respectively.

In this case, to find the concentration of Mg²⁺, we can rearrange the Nernst equation and solve for [Mg²⁺]. Plugging in the given values, we get: 2.527 V = 0 - (8.31 J/K/mol * 298 K)/(2 * 96500 C/mol) * ln([Mg²⁺]/(1.22 M)²). Solving for [Mg²⁺], we find that the concentration of Mg²⁺ is approximately 1.01 * 10^-4 M.