A voltaic cell is constructed based on the following reaction and initial concentrations:
Fe2+(0.0050 M) + Ag+(2.0 M) ↔ Fe3+(0.0050 M) + Ag(s)
Calculate [Fe2+] when the cell reaction reaches equilibrium.
To find the concentration of Fe2+ ions ([Fe2+]) when the cell reaction reaches equilibrium, we need to use the Nernst equation. The Nernst equation is derived from the standard cell potential equation for a half-cell reaction:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
- Ecell is the cell potential at equilibrium
- E°cell is the standard cell potential
- n is the number of electrons involved in the reaction
- Q is the reaction quotient.
In this case, the half-cell reaction involves the reduction of Ag+ ions to Ag(s), and the number of electrons involved is 1 (since Ag+ gains 1 electron to become Ag(s)).
Since the reaction has reached equilibrium, the cell potential is 0. Therefore, Ecell = 0 in the Nernst equation.
Ecell = 0 = E°cell - (0.0592/1) * log(Q)
To calculate Q, we need the concentrations of the species involved in the reaction.
In the given reaction, the concentration of Fe2+ is 0.0050 M.
The concentration of Fe3+ is also given as 0.0050 M.
The concentration of Ag+ is 2.0 M.
Since Ag(s) is a solid, its concentration remains constant and does not appear in the reaction quotient equation.
Q = [Fe3+]/[Fe2+][Ag+]
Plugging in the given concentrations, Q = (0.0050)/(0.0050 * 2.0)
Simplifying, Q = 0.5
Plugging this value for Q in the Nernst equation, we get:
0 = E°cell - (0.0592/1) * log(0.5)
Solving for E°cell, we find:
E°cell = (0.0592/1) * log(0.5)
Using a calculator, E°cell ≈ -0.0288 V
Now, we have the standard cell potential (E°cell) for the reaction.
To find [Fe2+] at equilibrium, we can use the Nernst equation and set Ecell to 0:
0 = -0.0288 - (0.0592/1) * log([Fe3+]/[Fe2+][Ag+])
Now, we substitute the known values:
0 = -0.0288 - (0.0592) * log(0.0050/[Fe2+]*2.0)
Simplifying, we have:
0.0592 * log(0.0050/[Fe2+]*2.0) = -0.0288
Taking the logarithm out of the equation:
log(0.0050/[Fe2+]*2.0) = -0.0288/0.0592
Using a calculator, the right side equals approximately -0.4872.
Now, taking the antilogarithm (10^x) of both sides:
0.0050/[Fe2+]*2.0 = 10^(-0.4872)
Simplifying:
0.010/[Fe2+] = 0.2997
Rearranging the equation:
[Fe2+] = 0.010/0.2997
Calculating, [Fe2+] ≈ 0.0334 M
Therefore, the concentration of Fe2+ ions at equilibrium is approximately 0.0334 M.