Answer:
The standard cell potential (E₀) for a Cu-Pb voltaic cell is +0.215V
The experiment is used to determine the standard electrode potential of the Cu-Pb Voltaic cell.
Explanation:
A Voltaic cell is one in which electric energy is produced as a result of the difference in electric potential between two electrodes in chemical solutions that are usually connected by a salt bridge.
In a voltaic cell, the anode electrode is where oxidation (loss of electrons) occurs while reduction (gain of electrons) occurs at the cathode electrode.
The standard cell potential (E₀) is the difference between the potential of the cathode and anode at standard conditions, that is, at standard temperature, pressure and concentration.
It is usually expressed as
E₀ = E₀cell (cathode) - E₀cell (anode)
From the given question, the half cell equations are :
Cu(s) - 2e ⇒ Cu²⁺. with oxidation reaction occuring at the anode
Pb²⁺ + 2e ⇒ Pb (s). with reduction reaction occuring at the cathode
Meaning that the solid copper electrode is oxidised by losing two electrons which are gained by the Pb²⁺ ions which are subsequently reduced to solid lead at the other electrode.
Each half cell reaction has a standard electric potential (recall that the standard electric potential is the potential at standard conditions, that is, at standard temperature, pressure and concentrations.
)
The overall voltaic cell equation is given as:
Cu(s) I Cu²⁺(aq) II Pb²⁺ (aq) I Pb(s)
The standard cell potential E₀ is obtained by
:
E₀ = E₀cell (cathode) - E₀cell (anode)
At standard conditions, E₀ (Cu(s) ⇒ Cu²⁺ + 2e) = - 0.340V
E₀ (Pb²⁺ + 2e ⇒ Pb(s) = - 0.125V
Therefore, the standard cell potential for a Cu-Pb voltaic cell is,
E₀ = -0.125V - (-0.340V)
= -0.125V + 0.340V
= +0.215V
