Chapter 21, Problem 12TE

What element behaves as the oxidizing agent in the following equation and what element behaves as the reducing agent?

${\text{Sn}}^{2+}+\text{2Ag}\to \text{Sn}+2{\text{Ag}}^{+}$

To determine

The elements that behave as an oxidizing agent and a reducing agent in the given equation.

${\text{Sn}}^{2+}+\text{2Ag}\to \text{Sn}+2{\text{Ag}}^{+}$

Answer to Problem 12TE

The tin ions ${\text{Sn}}^{2+}$ functions as an oxidizing agent and the silver atom $\text{Ag}$ functions as a reducing agent in the given equation.

Explanation of Solution

The electrons get donated in any oxidation reaction while in reduction reaction the electrons get accepted.

The oxidizing agent basically oxidizes another reactant/speciesand getting gets itself reduced. The reducing agent basically reduces another reactant/species while itself gets oxidized.

The given equation is shown below,

${\text{Sn}}^{2+}+\text{2Ag}\to \text{Sn}+2{\text{Ag}}^{+}$

The silver atoms, $\text{Ag}$ loses the electrons. So, it gets oxidized.And thus, itfunctions as a reducing agent.

The tin ions, ${\text{Sn}}^{2+}$ accepts the electrons. So, it gets reduced.And thus, itfunctions as an oxidizing agent.

Conclusion:

Therefore, the tin ions ${\text{Sn}}^{2+}$ functions as an oxidizing agent and the silver atom $\text{Ag}$ functions as a reducing agent in the given equation.