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
what is the pH of the solution in anode compartment of the following if the measured cell potential at 25oC is 0.28V
To calculate the E.M.F (voltage) of an electrochemical cell, you need to know the standard reduction potentials of the half-reactions involved in the cell and the stoichiometric coefficients of the balanced equation.
In the given redox reaction:
Zn(s) + Pb2+(aq) --> Zn2+(aq) + Pb(s)
The reduction half-reactions are:
Pb2+(aq) + 2 e- -> Pb(s) (Eº = -0.13V)
Zn2+(aq) + 2 e- -> Zn(s) (Eº = -0.76V)
To calculate the E.M.F (voltage) of the cell, you need to subtract the reduction potential of the anode (where oxidation occurs) from the reduction potential of the cathode (where reduction occurs).
In this case, the reduction potential of the anode (Pb to Pb2+) is -0.13V, and the reduction potential of the cathode (Zn2+ to Zn) is -0.76V.
Therefore, the E.M.F (voltage) of the cell is:
E.M.F = Reduction potential of cathode - Reduction potential of anode
E.M.F = (-0.76V) - (-0.13V)
E.M.F = -0.63V
Therefore, the E.M.F (voltage) of the cell is -0.63V.