In a fume cupboard molten lead(ii) bromide is electrolysed using graphite electrodes. Write half-equations for the reaction at the (i).cathode (ii). Anode (iii). Overall equation (states required).
(i) At the cathode:
The cathode is the negative electrode, and reduction occurs at the cathode during electrolysis. In this case, molten lead(II) bromide is being electrolyzed.
The half-equation for the reaction at the cathode can be written as follows:
Pb^2+(l) + 2e^- → Pb(l)
(ii) At the anode:
The anode is the positive electrode, and oxidation occurs at the anode during electrolysis. In this case, molten lead(II) bromide is being electrolyzed.
The half-equation for the reaction at the anode can be written as follows:
2Br^-(l) → Br2(g) + 2e^-
(iii) Overall equation:
To find the overall equation, we need to add the half-equations from (i) and (ii) together. However, as one of the half-equations involves the reduction of Pb^2+ ions to Pb metal, while the other involves the oxidation of Br^- ions to Br2 gas, we need to balance the number of electrons involved in both half-equations.
The balanced overall equation is as follows:
2Pb^2+(l) + 4Br^-(l) → 2Pb(l) + Br2(g)
Please note that the states (l) for liquid and (g) for gas are included to indicate the physical state of the substances involved in the reaction.
To write the half-equations and the overall equation for the electrolysis of molten lead(II) bromide (PbBr2) using graphite electrodes, we need to consider the oxidation and reduction reactions occurring at the anode and cathode, respectively.
(i) Cathode:
At the cathode, reduction occurs. In this case, lead(II) ions (Pb2+) from the molten lead(II) bromide are reduced to lead atoms (Pb).
The half-equation for the reaction at the cathode is:
Pb2+ + 2e- -> Pb
(ii) Anode:
At the anode, oxidation occurs. In this case, bromide ions (Br-) from the molten lead(II) bromide are oxidized to bromine gas (Br2).
The half-equation for the reaction at the anode is:
2Br- -> Br2 + 2e-
(iii) Overall Equation:
To obtain the overall equation, we need to combine the half-equations, ensuring that the number of electrons transferred is balanced.
Multiply the cathode half-equation by 2 to balance the number of electrons:
2Pb2+ + 4e- -> 2Pb
Now, add the balanced half-equations together to get the overall equation:
2Pb2+ + 4Br- -> 2Pb + Br2
The overall equation for the electrolysis of molten lead(II) bromide using graphite electrodes is:
2PbBr2 -> 2Pb + Br2 (solid state symbols are not required in the overall equation)
Note: The use of a fume cupboard is important as bromine gas produced during the electrolysis is toxic and must be safely vented. Additionally, as this experiment involves molten substances and high temperatures, proper precautions and safety measures must be followed.