Calculate the electrode potential of a copper half-call containing 0.05M Cu(NO3).
E^0 Cu2+/Cu = 0.337 V
College Chemistry - DrBob222, Tuesday, April 20, 2010 at 5:19pm
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?
As I stated in my response, substitute 2 for n and 0.05 M for Cu^+2. You substitute 1 for Cu metal (that is its standard state) and solve for E. E is the only unknown you have.
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To calculate the electrode potential of a copper half-cell containing 0.05M Cu(NO3), you can use the Nernst equation:
E = Eo - (0.0592/n)log[(Cu)/(Cu^2+)]
Where:
E is the electrode potential
Eo is the standard electrode potential (0.337 V in this case)
n is the number of moles of electrons transferred (2 for Cu2+ to Cu)
(Cu) is the concentration of Cu ions
(Cu^2+) is the concentration of Cu2+ ions
Since you want to calculate the electrode potential, you need to plug in the values for Eo, n, (Cu), and (Cu^2+), and then solve for E.
In this case, you are given that the concentration of Cu2+ is 0.05M. Therefore, you can plug in the values as follows:
E = 0.337 V - (0.0592/2)log[(Cu)/(0.05)]
Since you are not given the concentration of Cu, you cannot calculate the exact value of E without that information.
If you have the concentration of Cu, you can substitute the value into the equation and solve for E using logarithms and arithmetic.