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Grade 12Physical Chemistry

What would be the reduction potential of cadmium half cell having [Cd2+]= 0.01 at 250C ?

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9 Years agoGrade 12
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1 Answer

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ApprovedApproved Tutor Answer1 Year ago

To determine the reduction potential of a cadmium half-cell with a concentration of [Cd2+] = 0.01 M at 25°C, we can use the Nernst equation. This equation relates the concentration of ions in solution to the electrode potential, allowing us to calculate the potential under non-standard conditions.

The Nernst Equation

The Nernst equation is expressed as:

E = E° - (RT/nF) * ln(Q)

Where:

  • E = electrode potential under non-standard conditions
  • E° = standard electrode potential
  • R = universal gas constant (8.314 J/(mol·K))
  • T = temperature in Kelvin (25°C = 298 K)
  • n = number of moles of electrons transferred in the half-reaction
  • F = Faraday's constant (96485 C/mol)
  • Q = reaction quotient

Standard Electrode Potential for Cadmium

The standard reduction potential (E°) for the cadmium half-reaction:

Cd2+ + 2e- ⇌ Cd

is approximately -0.403 V. This value indicates that cadmium is a relatively weak oxidizing agent compared to other metals.

Calculating the Reaction Quotient (Q)

For the cadmium half-reaction, the reaction quotient Q is defined as:

Q = [Cd2+]

Given that [Cd2+] = 0.01 M, we have:

Q = 0.01

Plugging Values into the Nernst Equation

Now, we can substitute the values into the Nernst equation. First, we need to convert the temperature to Kelvin:

T = 25°C + 273.15 = 298.15 K

Next, we can calculate the potential:

E = -0.403 V - (8.314 J/(mol·K) * 298.15 K / (2 * 96485 C/mol)) * ln(0.01)

Calculating Each Component

Let's break down the calculation:

  • Calculate the term (RT/nF):
  • (8.314 * 298.15) / (2 * 96485) ≈ 0.0041 V

  • Now calculate ln(0.01):
  • ln(0.01) ≈ -4.605

  • Now substitute these values back into the equation:

E = -0.403 V - (0.0041 V * -4.605)

E ≈ -0.403 V + 0.0189 V

E ≈ -0.384 V

Final Result

The reduction potential of the cadmium half-cell at a concentration of [Cd2+] = 0.01 M at 25°C is approximately -0.384 V. This value indicates that the cadmium ion is still a relatively weak oxidizing agent, but the potential is slightly more favorable than under standard conditions due to the lower concentration of cadmium ions.