A concentation cell is consrtucted by o cadmium electrodes in 0.05M&o.1M cdso4 and the two solutions are connected by a salt bridge.construct the cell,cell representation and calculate EMF of the cell at 298K?

To construct a concentration cell using cadmium electrodes in two different concentrations of cadmium sulfate (CdSO4), we need to understand the setup and the calculations involved in determining the electromotive force (EMF) of the cell. Let's break this down step by step.

Cell Construction

A concentration cell consists of two half-cells that contain the same type of electrode but are immersed in solutions of different concentrations. In this case, we have:

• Left half-cell: Cadmium electrode in 0.05 M CdSO4
• Right half-cell: Cadmium electrode in 0.1 M CdSO4

To connect these two half-cells, we use a salt bridge, which allows the flow of ions while preventing the mixing of the two solutions. The overall cell can be represented as follows:

Cell Representation

The cell can be represented in shorthand notation as:

Cd(s) | CdSO4 (0.05 M) || CdSO4 (0.1 M) | Cd(s)

Here, the double vertical line (||) indicates the salt bridge separating the two half-cells.

Calculating the EMF of the Cell

The EMF of a concentration cell can be calculated using the Nernst equation, which is given by:

E = E° - (RT/nF) * ln(Q)

Where:

• E° is the standard electrode potential (for cadmium, E° = -0.403 V).
• R is the universal gas constant (8.314 J/(mol·K)).
• T is the temperature in Kelvin (298 K in this case).
• n is the number of moles of electrons transferred (for cadmium, n = 2).
• F is Faraday's constant (96485 C/mol).
• Q is the reaction quotient, which is the ratio of the concentrations of the products to the reactants.

Determining the Reaction Quotient (Q)

In a concentration cell, the reaction involves the transfer of cadmium ions from the higher concentration to the lower concentration. Therefore, the reaction quotient Q can be expressed as:

Q = [Cd²⁺] (lower concentration) / [Cd²⁺] (higher concentration)

Substituting the concentrations:

Q = 0.05 / 0.1 = 0.5

Plugging Values into the Nernst Equation

Now, we can substitute the values into the Nernst equation:

E = -0.403 V - (8.314 J/(mol·K) * 298 K / (2 * 96485 C/mol)) * ln(0.5)

Calculating the second term:

• First, calculate RT/nF:
• RT = 8.314 * 298 = 2477.572 J/mol
• nF = 2 * 96485 = 192970 C/mol
• RT/nF = 2477.572 / 192970 = 0.01284 V

Now, calculate ln(0.5):

ln(0.5) ≈ -0.693

Substituting this back into the equation:

E = -0.403 V - (0.01284 * -0.693)

E = -0.403 V + 0.0089 V ≈ -0.3941 V

Final Result

The EMF of the concentration cell at 298 K is approximately -0.394 V. This negative value indicates that the cell will produce a potential difference, driving the spontaneous flow of electrons from the higher concentration to the lower concentration, thus generating electrical energy.