
Identify the substances oxidized, reduced, oxidizing agent, and reducing agent for the given reaction:
\[HCHO(l)+2C{{u}^{2+}}(aq)+5O{{H}^{-}}\to C{{u}_{2}}O(s)+HCO{{O}^{-}}(aq)+3{{H}_{2}}O(l)\]
Answer
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Hint: The oxidizing agent and reducing agent in a reaction can be calculated by finding the oxidation state of the atoms in the reaction. The oxidized and reduced agents are the opposite of oxidizing and reducing agents.
Complete step by step answer:
The oxidation number of an element may be defined as the charge which an atom of the element has in its ion or appears to have when present in the combined state with other atoms. The oxidation number is also known as the oxidation state.
Oxidation may be defined as the chemical change in which there occurs an increase in the oxidation number of an atom or atoms.
The reduction may be defined as the chemical change in which there occurs a decrease in the oxidation number of an atom or atoms.
An oxidizing agent or an oxidant may be defined as a substance the oxidation number of whose atom decreases.
A reducing agent or a reluctant may be defined as a substance, the oxidation number of whose atom increases.
So, in the equation we have to find the oxidation number of every atom.
$\begin{align}
& +1\text{ 0 +1 -2 +2 -2 +1 +1 -2 +1 +2 -2 -2 +1 -2 } \\
& \text{H C H O(l) + 2C}{{\text{u}}^{\text{2+}}}\text{(aq) + 5O }{{\text{H}}^{\text{-}}}\to \text{C}{{\text{u}}_{\text{2}}}\text{ O(s) + H C O }{{\text{O}}^{\text{-}}}\text{(aq) + 3 H}{{\text{ }}_{\text{2 }}}\text{ O(l)} \\
\end{align}$
So, in the reaction oxidation number of carbon increases from 0 to +2 and the oxidation number of copper decreases from +2 to +1.
The oxidation number of copper decreases hence $2C{{u}^{2+}}$ is the oxidizing agent.
The oxidation number of carbon increases hence $HCHO$ is the reducing agent.
So, the $HCHO$ is oxidized and $2C{{u}^{2+}}$ is reduced.
Note: You may get confused that the substance which gets oxidized also acts as an oxidizing agent and the substance which gets reduced also acts as a reducing agent. But they are the opposite.
Complete step by step answer:
The oxidation number of an element may be defined as the charge which an atom of the element has in its ion or appears to have when present in the combined state with other atoms. The oxidation number is also known as the oxidation state.
Oxidation may be defined as the chemical change in which there occurs an increase in the oxidation number of an atom or atoms.
The reduction may be defined as the chemical change in which there occurs a decrease in the oxidation number of an atom or atoms.
An oxidizing agent or an oxidant may be defined as a substance the oxidation number of whose atom decreases.
A reducing agent or a reluctant may be defined as a substance, the oxidation number of whose atom increases.
So, in the equation we have to find the oxidation number of every atom.
$\begin{align}
& +1\text{ 0 +1 -2 +2 -2 +1 +1 -2 +1 +2 -2 -2 +1 -2 } \\
& \text{H C H O(l) + 2C}{{\text{u}}^{\text{2+}}}\text{(aq) + 5O }{{\text{H}}^{\text{-}}}\to \text{C}{{\text{u}}_{\text{2}}}\text{ O(s) + H C O }{{\text{O}}^{\text{-}}}\text{(aq) + 3 H}{{\text{ }}_{\text{2 }}}\text{ O(l)} \\
\end{align}$
So, in the reaction oxidation number of carbon increases from 0 to +2 and the oxidation number of copper decreases from +2 to +1.
The oxidation number of copper decreases hence $2C{{u}^{2+}}$ is the oxidizing agent.
The oxidation number of carbon increases hence $HCHO$ is the reducing agent.
So, the $HCHO$ is oxidized and $2C{{u}^{2+}}$ is reduced.
Note: You may get confused that the substance which gets oxidized also acts as an oxidizing agent and the substance which gets reduced also acts as a reducing agent. But they are the opposite.
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