A voltaic cell consists of a standard hydrogen electrode in one half-cell and a Cu/Cu2+ half-cell. Calculate [Cu2+] when E cell is 0.070 V.
NOTE::::::::::::::::::::ANSWER IN SCTIENTIFIC NOTATION PLEASE!!! thank you!
To calculate the concentration of Cu2+ ions ([Cu2+]) in the half-cell, you need to know the standard electrode potential (E°) for the Cu/Cu2+ half-reaction. The standard electrode potential for Cu/Cu2+ is +0.34 volts.
The formula used to calculate the cell potential (Ecell) is:
Ecell = E°Cu/Cu2+ - E°H+/H2
Given that Ecell is 0.070 volts and E°H+/H2 is 0.00 volts, you can rearrange the formula to solve for E°Cu/Cu2+:
E°Cu/Cu2+ = Ecell + E°H+/H2
= 0.070 + 0.00
= 0.070 volts
The Nernst equation relates the cell potential (Ecell) to the concentration of the species involved in the half-reactions. The Nernst equation for the Cu/Cu2+ half-reaction is:
Ecell = E°Cu/Cu2+ - (RT/nF) * ln([Cu2+])
Where:
R = gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
n = number of electrons transferred in the half-reaction (2 for Cu/Cu2+)
F = Faraday's constant (96,485 C/mol)
Assuming room temperature (25 degrees Celsius or 298 K), you can rearrange the Nernst equation to solve for [Cu2+]:
ln([Cu2+]) = (E°Cu/Cu2+ - Ecell) * (nF/RT)
[Cu2+] = e^((E°Cu/Cu2+ - Ecell) * (nF/RT))
Plugging in the values, you get:
[Cu2+] = e^((0.34 - 0.070) * (2 * 96485 / (8.314 * 298)))
Calculating this expression gives:
[Cu2+] ≈ 5.19e-4 M
Therefore, the concentration of Cu2+ ions is approximately 5.19 × 10^-4 M.
To calculate the concentration of Cu2+ in the half-cell, you can use the Nernst equation, which relates the cell potential (E cell) to the concentrations of the species involved.
The Nernst equation is given by:
E cell = E° cell - (RT/nF) * ln(Q)
Where:
- E cell is the cell potential
- E° cell is the standard cell potential
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- n is the number of electrons transferred in the balanced redox reaction
- F is the Faraday constant (96,485 C/mol)
- Q is the reaction quotient, which is the ratio of the concentrations of the products to the reactants.
In this case, the standard cell potential for the redox reaction involving Cu/Cu2+ and H+/H2 is given as 0.070 V. Therefore, E° cell = 0.070 V.
The reaction taking place in the half-cell involving Cu/Cu2+ is:
Cu(s) → Cu2+(aq) + 2e-
Since the reaction does not involve hydrogen, the reaction quotient Q will be equal to [Cu2+].
Plugging in the values in the Nernst equation, we get:
0.070 V = 0.070 V - (8.314 J/(mol·K) * T/(2 * 96485 C/mol)) * ln([Cu2+])
Simplifying the equation, we can cancel out the E° cell term:
0 = -(8.314 J/(mol·K) * T/(2 * 96485 C/mol)) * ln([Cu2+])
Again simplifying the equation, we get:
ln([Cu2+]) = 0
Taking the antilog of both sides, we get:
[Cu2+] = 1
Hence, the concentration of Cu2+ is 1 in scientific notation.
Cu ==> Cu^2+ + 2e Eo ox = ?
2H^+ + 2e ==> H2 Eo red = ?
-------------------
Cu + 2H^+ ==> Cu^2+ + H2 Eocell = ?
Then Ecell = Eocell - (0.05916/n)*logQ
where Q = (pH2)(Cu^2+)/(Cu)(H^+)^2
You know Ecell is 0.070v.
You know pH2 = 1 (it's standard H2 electrode).
You know (Cu) = 1 by definition.
You know Eocell from the above calculation which is the sum of the Eo ox and Eo red values.
Post your work if you get stuck.
Solve for (Cu^2+).