Chapter 19.2, Problem 31SSC

Interpretation Introduction

Interpretation:

The number of tin and gold ions are to be calculated to balance the electrons in the given redox reaction.

Concept introduction:

Oxidation half-reaction: The half part of redox reaction which represents oxidation and shows number of electrons lost.

Reduction half-reaction: The half part of redox reaction which represents reduction and shows number of electrons gained.

Answer to Problem 31SSC

Three ions of tin and two ions of gold are to required to have zero electrons leftover.

Explanation of Solution

Oxidation half reaction:

$S{n}^{+2}\to S{n}^{+4}+2e^{-}$

Reduction half reaction:

$A{u}^{+3}+3e^{-}\to Au$

To balance the reaction or to have zero electrons leftover, multiplying oxidation half reaction by 3 and reduction half reaction by 2 and adding

$3Sn+2A{u}^{+3}\to 3S{n}^{+2}+2Au$

Conclusion

Thus, three ions of tin and two ions of gold are to required to have zero electrons leftover.