Consider a cell at 299K:

line notation
Fe-Fe2+(1.39)--Cd2+(2.35)-Cd

Given the overall standard reduction potential of the cell is -0.038. What will the concentration of the Cd2+ solution be when the cell is dead?

To determine the concentration of Cd2+ in the solution when the cell is dead, we need to find the equilibrium condition of the cell reaction. The overall standard reduction potential of the cell, which is given as -0.038, can be used to determine the equilibrium condition.

First, let's understand the cell notation provided:
- Fe represents the anode (the electrode where oxidation occurs), and Fe2+ represents its species in the solution.
- Cd2+ represents the cathode (the electrode where reduction occurs), and Cd represents its species in the solution.

The line notation provided indicates the concentrations of the species involved in the cell reaction and their corresponding electrode potentials (measured against the standard hydrogen electrode, SHE).

Now, since the standard reduction potential of the cell is negative (-0.038), we can infer that the cell reaction is not spontaneous. In other words, when the cell is dead, the reaction has reached equilibrium, and the reduction potential of the cathode (Eº cell) will be zero.

To calculate the concentration of Cd2+ when the cell is dead, we can set up a Nernst equation using the given reduction potentials and the known value of Eº cell:

Eº cell = Eº cathode - Eº anode

To calculate Eº cathode and Eº anode, we can use the electrode potentials provided in the line notation. The reduction potential of Cd2+ (Eº cathode) is given as 2.35 V, and the reduction potential of Fe2+ (Eº anode) is given as 1.39 V.

Substituting the values into the equation, we have:

-0.038 V = 2.35 V - 1.39 V

Simplifying, we have:

-0.038 V = 0.96 V

Since Eº cathode - Eº anode = -0.038 V, the potential at equilibrium is -0.038 V.

Now, to determine the concentration of Cd2+ when the cell is dead, we can use the Nernst equation:

E = Eº - (RT / nF) * ln(Q)

Here, E is the potential at equilibrium (-0.038 V), R is the gas constant (8.314 J/(mol·K)), T is the temperature (299 K), n is the number of moles of electrons transferred (in this case, it is 2 electrons), F is the Faraday constant (96485 C/mol), and Q is the reaction quotient.

Since the cell is dead, the reaction is at equilibrium, so Q = 1. The Nernst equation then becomes:

-0.038 V = Eº - (8.314 J/(mol·K) * 299 K / (2 * 96485 C/mol) * ln(1)

Simplifying further, we have:

-0.038 V = Eº - 0

Therefore, the concentration of Cd2+ when the cell is dead does not depend on the equilibrium condition. The concentration of Cd2+ will remain the same as initially provided in the line notation.

To determine the concentration of Cd2+ when the cell is dead, we can use the Nernst equation. The Nernst equation relates the standard reduction potential of a cell and the concentrations of the species involved in the cell reaction.

The Nernst equation is given by:

Ecell = E°cell - (RT / nF) × ln(Q)

Where:
- Ecell is the cell potential under non-standard conditions
- E°cell is the standard reduction potential of the cell
- 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 cell reaction
- F is the Faraday constant (96485 C/mol)
- ln is the natural logarithm
- Q is the reaction quotient

In this case, the standard reduction potential of the cell is given as -0.038 V, the number of electrons transferred in the cell reaction can be determined by looking at the balanced reduction and oxidation half-equations:

Cd2+ + 2e- → Cd (reduction)
Fe2+ → Fe + 2e- (oxidation)

Thus, the number of electrons transferred (n) is 2.

Now, we substitute the given values into the Nernst equation and solve for Q:

-0.038 = 0.0592 / 2 × log(Q)

Simplifying the equation:

-0.038 = 0.0296 × log(Q)

Divide both sides by 0.0296:

-0.038 / 0.0296 = log(Q)

Now, calculate log(Q) using a calculator or logarithm table. The left side of the equation is -1.2838.

Therefore:

log(Q) = -1.2838

Now, take the antilog of both sides to find Q:

10^(log(Q)) = 10^(-1.2838)

Q = 0.056

The reaction quotient Q is equal to the product of the concentrations of the products divided by the concentrations of the reactants, each raised to the power of their stoichiometric coefficient.

In this case, the reaction quotient Q can be expressed as [Cd] / [Cd2+].

Since the overall concentration of Cd2+ when the cell is dead is being asked, we can assume that the concentration of Fe2+ is negligible compared to Cd2+. Therefore, [Cd2+] can be approximated as [Cd].

Hence, [Cd2+] ≈ [Cd] = 0.056

So, the concentration of the Cd2+ solution when the cell is dead is approximately 0.056 M.