Determine the unknown concentration of the ion in each of the following cells.
(a) Pb(s) | Pb2+(aq, ?) || Pb2+(aq, 0.10 mol·L-1) | Pb(s), E = +0.047 V.
(b) Pt(s) | Fe2+(aq, 1.0 mol·L-1) , Fe3+(aq, 0.10 mol·L-1) || Fe3+(aq, ?), Fe2+(aq, 0.0010 mol·L-1) | Pt(s), E = +0.17 V.
To determine the unknown concentration of the ion in each of the given cells, we need to use the Nernst equation. The Nernst equation relates the electrochemical potential of a half-reaction to the concentrations of the species involved. It is given by:
E = E° - (RT / nF) ln(Q)
Where:
- E is the cell potential (given in the problem).
- E° is the standard cell potential (known from literature).
- R is the gas constant (8.314 J·mol-1·K-1).
- T is the temperature in Kelvin.
- n is the number of electrons transferred in the balanced half-reaction.
- F is Faraday's constant (96,485 C·mol-1).
- Q is the reaction quotient, which is the ratio of the concentrations of products to reactants raised to their stoichiometric coefficients.
Now let's solve each part of the problem:
(a) Pb(s) | Pb2+(aq, ?) || Pb2+(aq, 0.10 mol·L-1) | Pb(s), E = +0.047 V.
In this cell, we have a solid lead electrode in contact with an unknown concentration of Pb2+ ions. The other half-cell has a known concentration of Pb2+ ions (0.10 mol·L-1) and a solid lead electrode as well. The cell potential is given as +0.047 V.
The Nernst equation for this cell is:
0.047 V = E° - (RT / 2F) ln(Q)
Since both half-cells have solid lead electrodes, the standard cell potential E° is zero. Therefore, the equation becomes:
0.047 V = -(RT / 2F) ln(Q)
To determine the concentration of Pb2+ ions (unknown in this case), we need to find Q. Q is the reaction quotient, which is the ratio of product to reactant concentration raised to their stoichiometric coefficients. The reaction is:
Pb(s) → Pb2+(aq) + 2e-
The stoichiometric coefficient for Pb2+(aq) is 1.
Therefore, Q = [Pb2+]/[Pb]. Since the concentration of Pb(s) will remain constant (solid lead electrode), we can replace [Pb] with 1.
0.047 V = -(RT / 2F) ln([Pb2+]/1)
Simplifying the equation, we have:
0.047 V = -(RT / 2F) ln([Pb2+])
Now we can rearrange the equation to solve for [Pb2+]:
[Pb2+] = exp((-0.047 V) * (2F / RT))
By plugging in the appropriate values for T (temperature), R (gas constant), and F (Faraday's constant), you can calculate the unknown concentration [Pb2+].
(b) Pt(s) | Fe2+(aq, 1.0 mol·L-1), Fe3+(aq, 0.10 mol·L-1) || Fe3+(aq, ?), Fe2+(aq, 0.0010 mol·L-1) | Pt(s), E = +0.17 V.
In this cell, we have a platinum electrode in contact with a solution containing Fe2+ (1.0 mol·L-1) and Fe3+ (0.10 mol·L-1). The other half-cell has a known concentration of Fe3+ (0.0010 mol·L-1) with Fe2+ (unknown concentration), and a platinum electrode. The cell potential is given as +0.17 V.
To determine the unknown concentration of Fe3+ ions, the Nernst equation can be applied:
0.17 V = E° - (RT / nF) ln(Q)
Since the E° and n values are not given in the question, we cannot solve for the exact concentration of Fe3+ ions without this information. The E° values for Fe2+ ↔ Fe3+ and Pb ↔ Pb2+ half-reactions are needed to calculate the cell potentials.
Please provide the standard cell potentials for the mentioned half-reactions to proceed with calculating the unknown concentrations.