Complete and balance the following equation: bro3−(aq)+n2h4(aq)→br2(l)+n2(g)(acidic solution)

Answer

BrO3- + 6H+ + 6e-= Br- + 3 H2O

N2H4 = N2 + 4 H+ + 4e-

2 BrO3- + 3 N2H4 = 2 Br- + 3 N2 + 6 H2O

3N2H4 + 2BrO3---------> 3N2 + 2Br- + 6H2O

BrO3- (aq) + N2H4(aq) --> Br2(l) + N2(g) , first let us balance
Br and N,

2 BrO3- (aq) + N2H4(aq) --> Br2(l) + N2(g) , now let us balance
O by adding H2O,

2BrO3- (aq) + N2H4 ---------> Br2 + N2 + 6H2O, now let us
balance H by adding H+ ions,

2BrO3- + N2H4 + 8H+ ---------> Br2 + N2 + 6H2O , finally charge
is balanced by adding electrons,

2 BrO3-(aq) + N2H4(aq) + 8H+(aq) + 6e- --> Br2(l) + N2(g) +
6H2O(l)

above eq is final balanced eq in acid medium