Can you please help me write the oxidation and reduction half reactions represented in each of these redox equations, and help me write the half reactions in net ionic form if they occur in aqueous solution.
1) PbO(s) + NH3(g) --> N2(g) + H2O(l) + Pb(s)
2) I2(s) + Na2S2O3(aq) --> Na2S2O4(aq) + NaI(aq)
3) Sn(s) +2HCl(aq) --> SnCl2(aq) + H2(g)
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Sure! Let's break down each redox equation and identify the oxidation and reduction half reactions.
1) PbO(s) + NH3(g) --> N2(g) + H2O(l) + Pb(s)
To find the oxidation and reduction half reactions, we need to determine the change in oxidation states for each element.
In PbO(s), lead (Pb) has an oxidation state of +2, and in Pb(s), it has an oxidation state of 0. Therefore, lead is reduced from +2 to 0. This gives us the reduction half reaction:
PbO(s) + 2H+(aq) + 2e- --> Pb(s) + H2O(l)
In NH3(g), nitrogen (N) has an oxidation state of -3, and in N2(g), it has an oxidation state of 0. Therefore, nitrogen is oxidized from -3 to 0. This gives us the oxidation half reaction:
NH3(g) --> 2H+(aq) + N2(g) + 3e-
To write the net ionic equations, we need to eliminate the spectator ions, which are the ions that do not change states during the reaction. In this case, there are no ions in the reaction, so there are no spectator ions. Therefore, the half reactions remain the same in net ionic form.
2) I2(s) + Na2S2O3(aq) --> Na2S2O4(aq) + NaI(aq)
In I2(s), iodine (I) has an oxidation state of 0, and in NaI(aq), it has an oxidation state of -1. Therefore, iodine is reduced from 0 to -1. This gives us the reduction half reaction:
I2(s) + 2e- --> 2I-(aq)
In Na2S2O3(aq), sulphur (S) has an oxidation state of +2, and in Na2S2O4(aq), it has an oxidation state of +4. Therefore, sulfur is oxidized from +2 to +4. This gives us the oxidation half reaction:
S2O32-(aq) --> S2O42-(aq) + 2e-
To write the net ionic equations, we need to eliminate the spectator ions. In this case, the spectator ion is Na+, which does not change states during the reaction. Therefore, the net ionic equations are as follows:
I2(s) + 2e- --> 2I-(aq)
S2O32-(aq) --> S2O42-(aq) + 2e-
3) Sn(s) + 2HCl(aq) --> SnCl2(aq) + H2(g)
In Sn(s), tin (Sn) has an oxidation state of 0, and in SnCl2(aq), it has an oxidation state of +2. Therefore, tin is oxidized from 0 to +2. This gives us the oxidation half reaction:
Sn(s) --> Sn2+(aq) + 2e-
In HCl(aq), chlorine (Cl) has an oxidation state of -1, and in H2(g), it has an oxidation state of 0. Therefore, chlorine is reduced from -1 to 0. This gives us the reduction half reaction:
2H+(aq) + 2e- --> H2(g)
To write the net ionic equations, we need to eliminate the spectator ions. In this case, the spectator ion is Cl-, which does not change states during the reaction. Therefore, the net ionic equations are as follows:
Sn(s) --> Sn2+(aq) + 2e-
2H+(aq) + 2e- --> H2(g)
I hope this helps! Let me know if you have any more questions.