The voltage of an electrochemical cell depends on the redox reaction occurring in the cell. Given a redox reaction below:
Zn(s) + Pb2+(aq) --> Zn2+(aq) + Pb(s)
Pb2+(aq) + 2 e--->Pb(s)Eº = -0.13V
Zn2+(aq) + 2 e--->Zn(s)Eº = -0.76V
Calculate the E.M.F (Voltage) of the cell
To calculate the E.M.F (voltage) of the cell, you can use the formula:
Ecell = Ecathode - Eanode
Where Ecathode is the reduction potential of the cathode (the reduction half-reaction) and Eanode is the reduction potential of the anode (the oxidation half-reaction).
Looking at the given redox reaction:
Zn(s) + Pb2+(aq) --> Zn2+(aq) + Pb(s)
- The reduction half-reaction is: Pb2+(aq) + 2 e- --> Pb(s) with a reduction potential of Eº = -0.13V.
- The oxidation half-reaction is: Zn(s) --> Zn2+(aq) + 2 e- with a reduction potential of Eº = -0.76V.
Now, we'll substitute these values into the formula:
Ecell = Ecathode - Eanode
Ecell = (-0.13V) - (-0.76V)
Simplifying the equation:
Ecell = 0.63V
Therefore, the E.M.F (voltage) of the cell is 0.63 volts.