Question

in the above redox reaction, use oxidation numbers to identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent.
xe + 3cu(oh)2 + oh− → 3cu + hxeo4− + 3h2o
name of the element oxidized:
formula of the oxidizing agent:
name of the element reduced:
formula of the reducing agent:
use the references to access important values if needed for this question.
3 item attempts remaining
submit answer
try another version

Explanation:

Step1: Determine oxidation numbers

In $Xe + 3Cu(OH)_2+OH^-\to3Cu + HXeO_4^-+3H_2O$, for $Xe$, its oxidation number changes from 0 in $Xe$ to +6 in $HXeO_4^-$. For $Cu$, its oxidation number changes from +2 in $Cu(OH)_2$ to 0 in $Cu$.

Step2: Identify oxidized and reduced elements

The element that loses electrons (increases in oxidation number) is oxidized. Here, $Xe$ is oxidized. The element that gains electrons (decreases in oxidation number) is reduced. Here, $Cu$ is reduced.

Step3: Identify oxidizing and reducing agents

The substance containing the reduced - element is the oxidizing agent. So, $Cu(OH)_2$ is the oxidizing agent. The substance containing the oxidized - element is the reducing agent. So, $Xe$ is the reducing agent.

Answer:

Name of the element oxidized: Xenon
Formula of the oxidizing agent: $Cu(OH)_2$
Name of the element reduced: Copper
Formula of the reducing agent: $Xe$