Calculate the copper(II)ion concentration at 25 °C in the cell Zn(s)| Zn²⁺(aq,1.0 M)|| Cu²⁺(aq)| Cu(s)if the measured cell potential is 1.06 V.The standard cell potential is 1.10 V. A) 0.21 M B) 1.0 M C) 0.045 M D) 1.0 M.. E) 1.0 M.

Question

Quizplus · Accepted Answer

First, write out the balanced redox reaction:

Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+(aq)

From the standard reduction potentials table, we know the standard reduction potential for the Cu2+/Cu half-reaction is +0.34 V, and for the Zn2+/Zn half-reaction is -0.76 V.

The standard cell potential, E°cell, is calculated by subtracting the reduction potential of the anode half-reaction from the reduction potential of the cathode half-reaction:

E°cell = E°Cu - E°Zn = +0.34 V - (-0.76 V) = +1.10 V

The measured cell potential, Ecell, is given as 1.06 V, which is lower than the standard cell potential. This means that the reaction is not at standard conditions, and the cell is not at equilibrium.

The Nernst equation can be used to calculate the non-standard cell potential:

Ecell = E°cell - (RT/nF) * ln(Q)

where R is the gas constant (8.314 J/mol*K), T is the temperature in Kelvin (25°C = 298 K), n is the number of electrons transferred in the reaction (2 in this case), F is the Faraday constant (96,485 C/mol), and Q is the reaction quotient.

The reaction quotient is calculated as the product of the concentrations of the species involved in the reaction, raised to their coefficients in the balanced equation:

Q = [Zn2+]/[Cu2+]

Since the copper electrode is the cathode, it is where reduction occurs, and Cu2+ is being reduced to Cu. Therefore, its concentration will decrease as the reaction proceeds. On the other hand, Zn is being oxidized at the anode, so its concentration will increase.

We assume that the solution volumes are constant, so the dilution effect can be ignored. Thus, the concentration of Cu2+ can be calculated as:

[Cu2+] = [Cu2+]° * exp[(E°cell - Ecell) * nF / (RT)]

where [Cu2+]° is the initial concentration of Cu2+ in the solution at standard conditions, which is given as 1.0 M.

Substituting the values:

[Cu2+] = 1.0 * exp[(1.10 - 1.06) * 2 * 96,485 / (8.314 * 298)] 
[Cu2+] = 0.045 M

Therefore, the copper(II) ion concentration at 25°C in the cell is 0.045 M, which is answer choice C.

Calculate the Copper(II)ion Concentration at 25 °C in the Cell