124k views
4 votes
Calculate the potential for the Cu²⁺ / Cu⁺ electrode at 25°c when the concentration of Cu⁺ ions is three times that of Cu²⁺ ions.

Cu²⁺ + e⁻ ----> Cu⁺

1 Answer

6 votes

Final answer:

To calculate the potential for the Cu²⁺ / Cu⁺ electrode at 25°C with the given ion concentrations, use the Nernst equation with Q set to 3, but the standard electrode potential E° must be known to complete the calculation.

Step-by-step explanation:

To calculate the potential for the Cu²⁺ / Cu⁺ electrode at 25°c when the concentration of Cu⁺ ions is three times that of Cu²⁺ ions, we can use the Nernst equation:

E = E° - (RT/nF) * ln(Q)

Where:

  • E = Electrode potential
  • E° = Standard electrode potential
  • R = Universal gas constant (8.314 J/(mol·K))
  • T = Temperature in Kelvin (298 K for 25°C)
  • n = Number of moles of electrons transferred in the reaction (n=1 for this half-reaction)
  • F = Faraday's constant (96485 C/mol)
  • Q = Reaction quotient = [Cu⁺]/[Cu²⁺]

Given that the concentration of Cu⁺ is three times that of Cu²⁺:

Q = [Cu⁺]/[Cu²⁺] = 3[Cu²⁺]/[Cu²⁺] = 3

We do not have the value for E° for the Cu²⁺/Cu⁺ couple, but if we did, we would proceed as follows:

Substitute R, T, n, F, and Q into the Nernst equation and solve for E:

E = E° - (8.314 * 298 / (1 * 96485)) * ln(3)

After calculating ln(3) and multiplying by the constants, we can subtract this value from the standard electrode potential to find the electrode potential E.

User Paul Facklam
by
8.4k points