Consider the following cell:
Pt|H2(g, 0.130 atm)|H (aq, ? M)||Ag (aq, 1.00 M)|Ag(s)
If the measured cell potential is 0.90 V at 25 °C and the standard reduction potential of the Ag /Ag half-reaction couple is 0.80 V, calculate the hydrogen ion concentration in the anode compartment.
Ecell = Eocell - (0.0592/n)log Q
Q = (H^+)^2(Ag)^2/pH2*(Ag^+)^2
To calculate the hydrogen ion concentration (H+) in the anode compartment, we can use the Nernst equation. The Nernst equation relates the cell potential to the concentration of reactants and products:
E = E° - (0.0592/n) * log(Q)
Where:
E = measured cell potential
E° = standard reduction potential
n = number of electrons transferred in the half-reaction
Q = reaction quotient
In this case, we are looking at the oxidation half-reaction of H2(g) to H+ (aq):
H2(g) → 2H+(aq) + 2e-
The standard reduction potential for this half-reaction is not given, but we can calculate it using the fact that the overall cell potential is 0.90 V and the standard reduction potential for the Ag/Ag couple is 0.80 V. The overall cell potential is the sum of the reduction and oxidation potentials:
0.90 V = reduction potential for Ag/Ag + oxidation potential for H2(g)
Solving for the oxidation potential of H2(g):
oxidation potential for H2(g) = 0.90 V - 0.80 V = 0.10 V
Now, we can use the Nernst equation to determine the H+ concentration in the anode compartment. The reaction quotient Q for the given cell is:
Q = [H+][Ag+]/[H2]
Since the concentration of Ag+ is given as 1.00 M, and there is no information given about the concentration of H2, we can assume it to be 1 (since it is in its standard state). Thus:
Q = [H+][Ag+]/1
Simplifying:
Q = [H+][Ag+]
Substituting the values into the Nernst equation:
0.10 V = 0.0592/2 * log([H+][1.00 M])
0.10 = 0.0296 * log([H+])
Dividing by 0.0296:
3.38 = log([H+])
Taking the antilog of both sides:
[H+] = 10^(3.38)
[H+] ≈ 2.64 x 10^3 M
Therefore, the hydrogen ion concentration in the anode compartment is approximately 2.64 x 10^3 M.