Electrode potential of cadmium is -0.4V and electrode potential of cadmium is -0.74V. [Cd²⁺] = 0.1M and [Cr³⁺] = 0.01M. Calculate the E_cell and E^°_cell.

To calculate the cell potential (E cell) and standard cell potential (E ° cell) for the given electrochemical reaction involving cadmium (Cd) and chromium (Cr), we can use the Nernst equation and the standard electrode potentials.

Given Data

• Electrode potential of Cd: E°(Cd) = -0.4 V
• Electrode potential of Cr: E°(Cr) = -0.74 V
• [Cd²⁺] = 0.1 M
• [Cr³⁺] = 0.01 M

Step 1: Calculate E ° cell

The standard cell potential (E ° cell) can be calculated using the formula:

E ° cell = E°(cathode) - E°(anode)

In this case, chromium is reduced (cathode) and cadmium is oxidized (anode):

E ° cell = E°(Cr) - E°(Cd)

Substituting the values:

E ° cell = (-0.74 V) - (-0.4 V) = -0.34 V

Step 2: Calculate E cell using the Nernst Equation

The Nernst equation is given by:

E cell = E ° cell - (RT/nF) * ln(Q)

Where:

• R = 8.314 J/(mol·K)
• T = Temperature in Kelvin (assume 298 K for standard conditions)
• n = number of moles of electrons transferred (for this reaction, n = 2)
• F = Faraday's constant (approximately 96485 C/mol)
• Q = reaction quotient

Step 3: Calculate Q

For the reaction:

Cd + Cr³⁺ → Cd²⁺ + Cr

The reaction quotient (Q) is:

Q = [Cd²⁺]/[Cr³⁺]

Substituting the concentrations:

Q = 0.1 / 0.01 = 10

Step 4: Substitute values into the Nernst Equation

Now, we can calculate E cell:

E cell = -0.34 V - (8.314 * 298 / (2 * 96485)) * ln(10)

Calculating the second term:

E cell = -0.34 V - (0.00412) * 2.303

E cell = -0.34 V - 0.00948 V

E cell ≈ -0.34948 V

Final Results

The calculated values are:

• E ° cell ≈ -0.34 V
• E cell ≈ -0.349 V