A voltaic cell is constructed based on the following reaction and initial concentrations:

Fe^2+,( 0.0055 M ) + Ag+,( 2.5 M ) ----> Fe^3+,( 0.0055 M ) + Ag(s)
<----
its a forward and reverse arrow

Calculate [Fe^2]+ when the cell reaction reaches equilibrium

can someone please explain how to do this?

See answer below:

http://www.jiskha.com/display.cgi?id=1238125870

To calculate the concentration of Fe^2+ when the cell reaction reaches equilibrium, you need to use the Nernst equation. This equation relates the concentration of reactants and products in an electrochemical cell to the cell potential.

The Nernst equation is as follows:
Ecell = E°cell - (RT/nF) * ln(Q)

Where:
- Ecell is the cell potential (which is 0 at equilibrium),
- E°cell is the standard cell potential,
- R is the gas constant (8.314 J/mol·K),
- T is the temperature in Kelvin,
- n is the number of electrons transferred in the reaction (in this case, 1),
- F is the Faraday constant (96,485 C/mol),
- Q is the reaction quotient, which is the ratio of products to reactants. In this case, Q = [Fe^3+]/[Ag+].

First, let's find the value of E°cell for this reaction. This can be obtained from standard reduction potentials of each half-cell reaction. The reduction potentials are as follows:
Fe^3+ + e^- → Fe^2+ E° = +0.77 V
Ag+ + e^- → Ag E° = +0.80 V

Add these two reduction potentials to get the standard cell potential (E°cell):
E°cell = +0.77 V - (+0.80 V) = -0.03 V

Since Ecell is 0 at equilibrium, the Nernst equation simplifies to:
0 = - (RT/nF) * ln(Q)

Now, let's plug in the known values and solve for Q:
0 = - (8.314 J/mol·K * T / (1 * 96,485 C/mol)) * ln(Q)

To determine the concentration of Fe^2+ at equilibrium, you will need the concentration of Fe^3+ and Ag+ at equilibrium. However, the initial concentrations are given. So the Q in the equation is Q = [Fe^3+ (at equilibrium)] / [Ag+ (at equilibrium)].

Since we want to find [Fe^2+] at equilibrium, we need to rearrange the Nernst equation to solve for Q first, and then compute [Fe^2+].

1. Rearrange the equation:
ln(Q) = 0

2. Calculate Q:
Q = e^0 = 1

3. Now, plug in Q into the equation:
0 = - (8.314 J/mol·K * T / (1 * 96,485 C/mol)) * ln(1)

Since ln(1) is 0, the whole equation becomes 0 = 0.

This means that at equilibrium, [Fe^2+] = 0.0055 M.

Therefore, when the cell reaction reaches equilibrium, the concentration of Fe^2+ will be 0.0055 M.