See answer below:
http://www.jiskha.com/display.cgi?id=1238125870
Fe^2+,( 0.0055 M ) + Ag+,( 2.5 M ) ----> Fe^3+,( 0.0055 M ) + Ag(s)
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its a forward and reverse arrow
Calculate [Fe^2]+ when the cell reaction reaches equilibrium
can someone please explain how to do this?
http://www.jiskha.com/display.cgi?id=1238125870
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.