Oxidation and Reduction

Question 1

Electrolytic M part

Answer 1

The oxidation occurs at the anode left side electrode and is positive and is attracted to the colourless Br- ions that is oxidised to Br2 liquid an orange liquid.

Electrons are lost here and travel through the wire to the cathode.

The reduction occurs at the cathode right-side electrode and is negative and is attracted to the colourless Sn2+ ions that gain the electron lost by B- ions and are reduced to an Sn solid metal. This is observed by a grey deposit forming around the cathode where the solution is colourless as Br- and Sn2+ are also colourless.

Question 2

Electrochemical M Part

Answer 2

The oxidation occurs at the anode left electrode, At the anode a grey solid Zn metal (Zn) losses mass as Zn is oxidised into Zn2+ ions. Zn2+ ion is added to the solution due to this the solution will remain colourless.

Electrons are lost here and travel through the wire to the cathode.

Reduction occurs at the cathode on the right-hand side. The cathode is made of Sn metal. The cathode will gain mass as a grey deposit of Sn metal is formed as the Sn2+ ion in the solution gain electron and are reduced into Sn metal. As Sn2+ ions are being used up the solution will remain colourless.

Question 3

Electrochemical and Electrolytic E part

Answer 3

In the electrolytic cell, the reduction potential for Pb2+/Pb couple E°(Pb2+(aq)/Pb(s))= -0.13V cell potential is lower than the Cl2/Cl- couple E°(Cl-(s)/Cl2(g))= +1.36 meaning that Pb2+ ion is a weaker oxidant than the Cl2, as Pb2+ isn’t able to fully oxidise the Cl- ion to Cl2 gas without the help of an external power source. The cell potential for electrolytic reaction is E cell°(-1.49V), thus is not spontaneous . A voltage of more than 1.49 V must be applied using a battery or external power source to provide the energy required for the Cl- ions to give up their extra
electron to the Pb2+ ions

In the electrochemical cell, the reduction potential for Pb2+/Pb couple E°(Pb2+(aq)/Pb(s))= -0.13V cell potential is higher than the Fe2+/Fe couple E°(Cl-(s)/Cl2(g))= +1.36, meaning that Pb2+ is a powerful oxidant than the Fe2+ and is able to oxide the fe(s) to Fe2+(aq) without the help of an external power source. The cell potential for the electrochemical reaction is E cell°(0.34), thus is spontaneous. A voltage of more than 0.34V is not required a battery or external power source to provide the energy required for the Fe2+ ions to give up their extra electron to the Pb2+ ions

Question 4

both E part conclusion

Answer 4

In both experiments Pb2+ is acting as the oxidant, being reduced to form Pb metal. In the electrochemical cell the Pb2+ is a powerful enough oxidant to oxidise the Fe(s) to Fe2+(aq) and allow the reaction to occur and produce Pb metal without the supply of energy from an external power source. Whereas, in the electrolytic cell, Pb2+ is not a powerful enough oxidant to oxidise the Cl-(aq) to Cl2(g) and produce the Pb metal without supplying electrical energy from an external power source.