Chapter 19, Problem 43A

Interpretation Introduction

Interpretation:

The reducing agent in given reaction is to be determined.

Concept introduction:

Reducing agent: Reducing agent is a species which helps for reduction.

The substance which undergoes increase in oxidation number is reducing agent.

Answer to Problem 43A

$Sn$ is reducing agent.

Explanation of Solution

The given equation is:

$8H^{+}+Sn+6C{l}^{-}+4N{O}_3{}^{-}\to SnC{l}_6{}^{-2}+4N{O}_2+4H_{2}O$

Oxidation number of various atoms is:

$8H^{+}+\stackrel{0}{Sn}+6C{l}^{-}+4\stackrel{+5}{N}{O}_3{}^{-}\to \stackrel{+4}{Sn}C{l}_6{}^{-2}+4\stackrel{+4}{N}{O}_2+4H_{2}O$

Oxidation no. of tin is increasing hence it is reducing agent.

Conclusion

Oxidation no. of tin is increasing hence it is reducing agent.