The following electrochemical cell has a potential of +0.326 V at 25 °C. Pt | H₂(g,1.00 atm) | H⁺(aq,1.00 M) || Cl^-(aq) | AgCl(s) | Ag
The standard reduction potential,E°,of AgCl(s) is +0.222 V.Calculate the Cl^-(aq) ion concentration.

Question

Quizplus · Accepted Answer

We can use the Nernst equation to calculate the concentration of Cl- ions:
E = E° - (RT/nF) * ln(Q)
where R is the gas constant, T is temperature in Kelvin, n is the number of electrons transferred, F is Faraday's constant, and ln(Q) is the natural logarithm of the reaction quotient.
Since the cell is at standard pressure and the temperature is 25 °C (298 K), we can simplify the equation to:
E = E° - (0.0592/n) * log(Q)
where log is the base-10 logarithm.
In this case, n is 1 because 1 electron is transferred in the reduction of AgCl(s) to Ag(s).
Plugging in the values, we get:
0.326 V = 0.222 V - (0.0592/n) * log(Q)
Solving for log(Q), we get:
log(Q) = -(0.326 V - 0.222 V) * n / 0.0592 = -1.7627 n
Since the reaction is:
AgCl(s) + e- → Ag(s) + Cl-
the reaction quotient is:
Q = [Ag+] / [Cl-]
Since the AgCl(s) is a solid, its concentration is constant and doesn't affect Q, so we can write:
Q = [Cl-]
Substituting this into the expression for log(Q), we get:
log([Cl-]) = -1.7627
Taking the antilogarithm (i.e. 10 raised to the power of), we get the concentration of Cl- ions:
[Cl-] = 10^(-1.7627) = 0.018 M
Therefore, the best choice is C.

The Following Electrochemical Cell Has a Potential of +0