Define oxidation and reduction. Balance the following redox reaction by ` oxidation number method or ion electrons method.` I_(2)+OH^(-)rarr I^(-)+10_(3)^(-)+H_(2)O

HNO₃ + H₃AsO₃(aq) → NO(g) + H₃AsO₄(aq) + H₂O(l)

Identify the oxidation number of every atom.

Left hand side: H= +1; N= +5; O = -2; As = +3

Right hand side: N = +2; O = -2; H = +1; As = +5

Determine the change in oxidation number for each atom that changes.

N: +5 → +2; Change = -3

As: +3 → +5; Change = +2

Make the total increase in oxidation number equal to the total decrease in oxidation number.

We need 2 atoms of N for every 3 atoms of As. This gives us total changes of -6 and +6.

Place these numbers as coefficients in front of the formulas containing those atoms.

2HNO₃ + 3H₃AsO₃(aq) → 2NO(g) + 3H₃AsO₄(aq) + H₂O(l)

Balance all remaining atoms other than H and O.

Done.

Balance O.

Done.

Balance H.

Done.

Explanation:

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8