Chapter 20, Problem 1SAQ

Balance the redox reaction equation (occurring in acidic solution) and choose the correct coefficients for each reactant and product.

   $\_\_{\text{VO}}_{\text{2}}{}^{+}\left(aq\right)+\text{ Sn}\left(s\right)+\_\_{\text{H}}^{+}\left(aq\right)\to \_\_{\text{VO}}^{\text{2}+}\left(aq\right)+{\text{ Sn}}^{\text{2}+}\left(aq\right)+\_\_{\text{H}}_{\text{2}}\text{O }\left(l\right)$

1. $\text{1},\text{1},\text{2}\to \text{1},\text{1},\text{1}$
2. $2,\text{1},\text{2}\to 2,\text{1},2$
3. $2,\text{1},4\to 2,\text{1},2$
4. $\text{2},\text{1},\text{2}\to 2,\text{1},\text{1}$

Interpretation Introduction

Interpretation: The given redox reaction occurs in acidic solution should be balanced and identifythe correct coefficients for each reactant and product.

  ${\text{\_VO}}_{\text{2}}^{\text{+}}\left(\text{aq}\right)\text{+Sn}\left(\text{s}\right)+{\text{\_H}}^{\text{+}}\to \_{\text{VO}}^{\text{2+}}\left(\text{aq}\right){\text{+Sn}}^{\text{2+}}\left(\text{aq}\right)+\_{\text{H}}_{\text{2}}\text{O(l)}$

Concept introduction:

The redox reaction is the reaction in which transfer of electrons between chemical species occurs which results in simultaneous oxidation and reduction process.

To balance the chemical reaction, the number of atoms present in reactant side must be equal to the number of atoms in product side.

Correct answer: Option (c) is correct i.e. $2,1,4\to 2,1,2$

Explanation of Solution

Reason for correct answer:

• Unbalanced redox reaction is:

  ${\text{\_VO}}_{\text{2}}^{\text{+}}\left(\text{aq}\right)\text{+Sn}\left(\text{s}\right)\to \_{\text{VO}}^{\text{2+}}\left(\text{aq}\right){\text{+Sn}}^{\text{2+}}\left(\text{aq}\right)$

• Half-reactions are:

  $\begin{array}{l}{\text{Reduction: VO}}_{\text{2}}^{\text{+}}\left(\text{aq}\right)\to {\text{VO}}^{\text{2+}}\left(\text{aq}\right)\\ \text{Oxidation: Sn}\left(\text{s}\right)\to {\text{Sn}}^{\text{2+}}\left(\text{aq}\right)\end{array}$

• Balancing the half-reactions:
Oxidation half-reaction:

  $\text{Sn}\to {\text{Sn}}^{\text{2+}}\text{+}{\text{2e}}^{-}$

Reduction half-reaction:

  ${\text{VO}}_{\text{2}}^{\text{+}}\to {\text{VO}}^{\text{2+}}$

Balancing O:

${\text{VO}}_{\text{2}}^{\text{+}}\to {\text{VO}}^{\text{2+}}+{\text{H}}_{\text{2}}\text{O}$

Balancing H:

${\text{VO}}_{\text{2}}^{\text{+}}+{\text{2H}}^{\text{+}}\to {\text{VO}}^{\text{2+}}+{\text{H}}_{\text{2}}\text{O}$

Balancing charge:

${\text{VO}}_{\text{2}}^{\text{+}}+{\text{2H}}^{\text{+}}+{\text{e}}^{\text{-}}\to {\text{VO}}^{\text{2+}}+{\text{H}}_{\text{2}}\text{O}$

• Balancing the number of electrons in both oxidation and reduction half-reactions:

  $\text{Sn}\to {\text{Sn}}^{\text{2+}}\text{+}{\text{2e}}^{-}$

  ${\text{(VO}}_{\text{2}}^{\text{+}}+{\text{2H}}^{\text{+}}+{\text{e}}^{\text{-}}\to {\text{VO}}^{\text{2+}}+{\text{H}}_{\text{2}}\text{O)}\times \text{2}$

• Adding both the half-reactions:

  $\begin{array}{l}\text{Sn}\to {\text{Sn}}^{\text{2+}}\text{+}{\text{2e}}^{-}\\ \underset{\_}{{\text{2VO}}_{\text{2}}^{\text{+}}+4{\text{H}}^{\text{+}}+2{\text{e}}^{\text{-}}\to 2{\text{VO}}^{\text{2+}}+2{\text{H}}_{\text{2}}\text{O}}\\ \text{Sn}\text{+}{\text{2VO}}_{\text{2}}^{\text{+}}+4{\text{H}}^{\text{+}}\to {\text{Sn}}^{\text{2+}}\text{+}2{\text{VO}}^{\text{2+}}+2{\text{H}}_{\text{2}}\text{O}\end{array}$

Thus, the balanced equation is:

  ${\text{2VO}}_{\text{2}}^{\text{+}}+\text{Sn}+4{\text{H}}^{\text{+}}\to 2{\text{VO}}^{\text{2+}}+{\text{Sn}}^{\text{2+}}\text{+}2{\text{H}}_{\text{2}}\text{O}$

Therefore, the correct option is (c) that is $2,1,4\to 2,1,2$

Option (a), (b) and (d) is incorrect answer