Question

Balance the equation, by ion electron method. 

As₂S₃(s) + NO₃^–(aq) + H⁺(aq) → AsO₄^3-(aq) + S(s) + NO(g) + H₂(l)

Answer 1

Step-1: To identify the atoms whose oxidation numbers have undergone a change. Writing the oxidation number of each atom above its symbol,

we have

Step-2:

Oxidation:

[As, from +3 to +5 i.e., 2 unit change, S from to O also 2 in change]

Reduction:

Step-3: To balance the oxidation half Eq. (i)

(a) Balance all the atoms other than H and O. Multiply AsO₄^3- by 2 and S by 3 on RHS of Eq. (i) we have,

(c) Balance charge by adding H⁺ ions. The total charge on RHS of Eq. (iv) is -16 and zero the LHS therefore, add 16H⁺ to RHS of Eq. (iv) we have,

(d) Balance O atoms by adding H₂O molecules.

The H-atoms get automatically balanced. Thus Eq. (vi) represents the balanced oxidation half equation.

Step-4: To balance the reduction half Eq. (ii)

(a) Balance oxidation number by adding electrons.

(b) Balance charge by adding H⁺ ions.

(c) Balance H⁺ by adding H₂O

Thus, Eq. (ix) represents the balanced reduction half equation Eq. (ix) by 10 and Eq, (vi) by 3 and (ix) 10 + (vi) x 3

Step-5: