Galvanic cells how to work out the cell potential using reduction potentials?
A galvanic cell is set up with a copper electrode in contact with CuSO4(aq) and a lead electrode in contact with Pb(NO3)2(aq), respectively, at 25ºC.
The standard reduction potentials are:
Pb2+ + 2 e- → Pb Eº= -0.13 V
Cu2+ + 2 e- → Cu Eº= +0.34 V
If the concentrations of Pb2+ and Cu2+ are each 1.0 M, what is the cell potential in Volt?
The standard reduction potentials are:
Pb2+ + 2 e- → Pb Eº= -0.13 V
Cu2+ + 2 e- → Cu Eº= +0.34 V
If the concentrations of Pb2+ and Cu2+ are each 1.0 M, what is the cell potential in Volt?
Best Answer - Chosen by Voters
First you need to identify the anode and cathode of the cell.
The anode is the half-cell with the standard reduction potential which is more negative. So, Pb half cell is the anode and oxidation will occur at anode.
The copper half cell which is more positive in SRP is the cathode.
Since the concentrations of Pb2+ and Cu2+ are 1.0M, use this formula to calculate the cell potential:
Cell potential = E(cathode) - E(anode)
.................. = (+0.34) - ( - 0.13)
.................. = 0,47 V
The anode is the half-cell with the standard reduction potential which is more negative. So, Pb half cell is the anode and oxidation will occur at anode.
The copper half cell which is more positive in SRP is the cathode.
Since the concentrations of Pb2+ and Cu2+ are 1.0M, use this formula to calculate the cell potential:
Cell potential = E(cathode) - E(anode)
.................. = (+0.34) - ( - 0.13)
.................. = 0,47 V
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