Question

Balance redox reaction by oxidation number and ion electron method

Answer 1

Ion electron Method

Sn^o + \(\overset{+5}{NO}\)₃^- + H⁺ \(\rightarrow\) Sn⁺² + \(\overset{-3}{NH}\)₄⁺ + H₂O

first of all separating half cell reaction

Sn \(\rightarrow\) Sn⁺² +2e^- (oxidation)....(i)

No₃^- + 8e^- \(\rightarrow\) NH₄⁺ (reduction)....(ii)

In equation (ii), adding 3H₂o in the right side (for balancing oxygen)

NO₃^- + 8e^-\(\rightarrow\) NH₄⁺ + 3H₂O...(iii)

Now, balancing atom, and charge adding 10H⁺ in the left side in equation (iii) we got

NO₃^- + 8e^- + 10H⁺ \(\rightarrow\) NH₄^{\(\bigoplus\)} + 3H₂O....(iv)

Now, multiply by '5' in equation (i) and then adding in equation (iv) we got

5Sn + NO₃^- + 10H⁺ \(\rightarrow\) 5Sn⁺² + NH₄⁺ + 3H₂O....(v)

equation (v) is a balance equation 

Ion electron Method

\(\overset{0}{AS}\) + \(\overset{+5}{NO}\) ₃^- + H⁺ \(\rightarrow\) A\(\overset{+5}{SO}\)₄^-3 + \(\overset{+4}{NO}\)₂ + H₂O

separating half cell reaction

\(\overset{+5}{NO}\)₃^- + e^- \(\rightarrow\) \(\overset{+4}{NO}\)₂....(i)

\(\overset{0}{AS}\) \(\rightarrow\) A \(\overset{+5}{SO}\)₄^-3 + 5e^-  ...(ii)

For balancing oxygen, adding (i) we get

NO₃^- + e^- \(\rightarrow\) NO₂ + H₂O....(iii)

For balancing Hydrogen and charge , adding 2H⁺ in the left side in the  equation (iii) we got

NO₃^- + 2H⁺ + e^- \(\rightarrow\) NO₂ + H₂O....(iv)

For equation (ii) 

For balancing oxygen, adding 4H₂O in left side, ion equation (ii) we got

AS^- + 4H₂O\(\rightarrow\) ASO₄^-3 + 5e^-....... (v)

For balancing Hydrogen, and change adding OH⁺ in the right side, in equation (v)

we got,

AS +  4H₂O\(\rightarrow\) ASO₄^-3 + 8H⁺ + 5e^-...(vi)

Now equation (iv) Multiply by '5' then adding equation (vi) we got

AS + 5NO₃^- + 2H⁺ \(\rightarrow\) ASO₄^-3 + 5NO₂ + H₂O.....(vii)

equation (iv) is balance equation