Calculate the electrode potential of a copper half-call containing 0.05M Cu(NO3).
E^0 Cu2+/Cu = 0.337 V
E = Eo - (0.0592/n)log[(Cu)/(Cu^+2)]
Substitute 2 for n and 0.05 M for Cu^+2 and 0.337 for Eo. Solve.
Am I solving for Cu or do I plug a value in for that also?
Answered above.
To calculate the electrode potential of a copper half-cell, we need to use the Nernst equation. The Nernst equation is given by:
E = E^0 - (RT/nF) * ln(Q)
Where:
E = the electrode potential
E^0 = the standard electrode potential
R = the gas constant (8.314 J/(mol·K))
T = the temperature in Kelvin
n = the number of electrons involved in the redox reaction
F = Faraday's constant (96,485 C/mol)
ln = the natural logarithm
Q = the reaction quotient
In this case, we have a copper half-cell containing Cu2+ ions. The reaction involved is:
Cu2+ + 2e^- -> Cu
The number of electrons involved (n) is 2.
We are given the standard electrode potential (E^0) for the Cu2+/Cu half-reaction, which is 0.337 V.
Finally, we are given the concentration of Cu2+ ions (0.05 M), which will be used to calculate the reaction quotient (Q).
Let's assume the temperature is 25°C, which is 298 K.
Now we can plug the values into the Nernst equation:
E = E^0 - (RT/nF) * ln(Q)
E = 0.337 V - [(8.314 J/(mol·K)) * (298 K) / (2 * 96,485 C/mol)] * ln(0.05)
Simplifying the equation should give us the electrode potential (E) for the copper half-cell.