Parta)Many oxidation-reduction reactions con be balanced byinspection, Try to balance the follaning reactions byinspection. In each reaction, identify the substancereducad and the substance oxidized.a) Al6) +HC lee)->AICI,69)+H,G)b)CHG)*So>CS,)+H,SG)OXPant b)Balance each of the following oxidation-reduction reactionsby using the oxidation states methad2+

Since you have posted questions with multiple subparts as per guidelines we can answer only first three. If you want remaining please repost. Thankyou Given : a) Al(s) + HCl(aq)-->AlCl3 (aq) + H2(g) b) CH4 (g) + S(s) ---> CS2 (l) + H2S(g) C) C3H8(g) + O2 (g) -->CO2 (g) + H2O(l)Briefing: a) Al(s) + HCl(aq)-->AlCl3 (aq) + H2(g) Since number of Cl on L.H.S is 1 and on R.H.S it is 3, so multiply by 3 on L.H.S Al(s) + 3HCl(aq)-->AlCl3 (aq) + H2(g) Now number of hydrogen atoms are different on both sides. Al(s) + 3HCl(aq)-->AlCl3 (aq) + 3/2H2(g) Now , 2Al(s) + 6HCl(aq)--> 2AlCl3 (aq) +3 H2(g) Since Al is in 0 oxidation state and on the product side Al is in +3 oxidation state and hence , oxidation is taking place. And hydrogen on L.H.S is in +1 oxidation state and on R.H.S it is in 0 oxidation state and hence , reduction is taking place. b) CH4 (g) + S(s) ---> CS2 (l) + H2S(g) It should be S8 not S(s) CH4 (g) + S8 (s) ---> CS2 (l) + H2S(g) Number of S on both sides are different , multiply CS2 with 2 and water with 4 CH4 (g) + S8 (s) ---> 2CS2 (l) +4 H2S(g) Now , number of carbon atom on L.H.S are different. 2CH4 (g) + S8 (s) ---> 2CS2 (l) +4 H2S(g) Sulphur on R.H.S is in -2 oxidation state and on L.H.S it is in 0 oxidation state and hence , reduction is taking place. Carbon is in - 4 oxidation state on L.H.S and +4 on R.H.S and hence , oxidation is taking place. C) C3H8(g) + O2 (g) -->CO2 (g) + H2O(l) Number of carbon atoms on L.H.S is different and hence multiply with 3 on R.H.S C3H8(g) + O2 (g) -->3CO2 (g) + H2O(l) Number of hydrogen atoms are less on R.H.S and hence multiply by 4 C3H8(g) + O2 (g) -->3CO2 (g) + 4 H2O(l) Now , number of oxygen atoms are different and hence , multiply by 5 on R.H.S C3H8(g) + 5O2 (g) -->3CO2 (g) + 4 H2O(l) Oxygen on R.H.S is in -2 oxidation state and on L.H.S it is in 0 oxidation state and hence , reduction is taking place. Carbon is in 8/3 on L.H.S and + 4 on R.H.S and hence oxidation is taking place.