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).

Question

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).

Answer 1

PbBr2 ==> Pb^2+ + 2Br^- are the ions you have.

Pb^2+ + 2e ==> Pb a the cathode (you can add the states).
2Br^- ==> Br2 + 2e at the anode (again you add the states).

Add the two half equations to obtain the overall equation.

Answer 2

(i) At the cathode:

The molten lead(II) bromide (PbBr2) will be reduced at the cathode electrode (negative electrode) during electrolysis. The half-equation for the reduction reaction is:

Pb2+ + 2e- -> Pb(s)

(ii) At the anode:
The bromide ions (Br-) from the molten lead(II) bromide will be oxidized at the anode electrode (positive electrode) during electrolysis. The half-equation for the oxidation reaction is:

2Br- -> Br2(g) + 2e-

(iii) Overall equation:
When you combine the reduction and oxidation half-equations, you get the overall equation for the electrolysis of molten lead(II) bromide:

PbBr2(l) -> Pb(s) + Br2(g)

Note: (l) indicates liquid state, (s) indicates solid state, and (g) indicates gaseous state.

Answer 3

To write the half-equations for the electrolysis of molten lead(II) bromide using graphite electrodes, we need to understand the ionic compounds involved and the principles of electrolysis.

First, let's break down the compound: lead(II) bromide (PbBr2) consists of Pb2+ cations and Br- anions.

During electrolysis, the positive Pb2+ cations migrate towards the negative electrode (cathode), where reduction occurs. The negative Br- anions migrate towards the positive electrode (anode), where oxidation occurs.

(i) Cathode half-equation:
At the cathode, the Pb2+ cations gain electrons to form solid lead (Pb). The half-equation for the cathode reaction is:
Pb2+ + 2e- -> Pb(s)

(ii) Anode half-equation:
At the anode, the Br- anions lose electrons to form bromine gas (Br2). The half-equation for the anode reaction is:
2Br- -> Br2(g) + 2e-

(iii) Overall equation:
By combining the cathode and anode half-equations, we get the overall equation for the electrolysis of molten lead(II) bromide:
Pb2+ + 2Br- -> Pb(s) + Br2(g)

Note: Please note that in this reaction, molten lead(II) bromide (PbBr2) is used. The state of the reactants and products is indicated as (s) for solid and (g) for gas.