some chemical equation type and mechanism

Question

this reaction is copy from my exercise book but i have never seen ........
i want to know how the product come from ....and follow which type of reaction

1. BrO3-(aq) + I- (aq) + H+(aq) ----------->

2.MnO2(s) + HCl (conc) --------------------->

3.KMn04(s)+ HCl (conc)---------------------->

Answer 1

1.
BrO3-(aq) + 6H+(aq) + 6e- → Br-(aq) + 3H2O(l) ... (*)
2I-(aq) → I2(aq) + 2e- ... (**)
(*) + 3x(**)
BrO3-(aq) + 6H+(aq) + 6I-(aq) →Br-(aq) + 3H2O(l) + 3I2(aq)

2.
MnO­2(aq) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l) ... (*)
2Cl-(aq) → Cl2(g) + 2e- ... (**)
(*) + (**)
MnO­2(aq) + 4H+(aq) + 2Cl-(aq) →Mn2+(aq) + 2H2O(l) + Cl2(g)

3.
KMnO4(s) + 8H+(aq) + 5e- → K+(aq) + Mn2+(aq) + 4H2O(l) ... (*)
2Cl-(aq) → Cl2(g) + 2e- ... (**)
2x(*) + 5*(**)
2KMnO4(s) + 16H+(aq) + 10Cl-(aq) → 2K+(aq) + 2Mn2+(aq) + 8H2O(l) + 5Cl2(g)
Assume that KMnO4 dissolves before reaction, the equation becomes :
2MnO4-(s) + 16H+(aq) + 10Cl-(aq) → 2Mn2+(aq) + 8H2O(l) + 5Cl2(g)

Answer 2

For no.2 the products are MnCl2(aq), Cl2(g) and H2O(l). Pls balance the equation yourself.

No.2 is a redox reaction, in which the oxidizing agent MnO2 is reduced to Mn2+ and the reducing agent Cl- is oxidized to Cl2.

For no.3 the products are KCl (aq), MnCl2(aq), Cl2(g) and H2O(l). Pls balance the equation yourself.

No.3 is also a redox reaction, in which the oxidizing agent KMnO4 is reduced to Mn2+ and the reducing agent Cl- is oxidized to Cl2.