Identify the substances that are oxidised and the substances that are reduced in the following reaction:
CuO(s) + H2(g) → Cu(s) + H2O(l)
Answer & explanation
Correct answer: option 1
The correct answer is option 1. \(CuO\) is reduced and \(H_2\) gets oxidized.
To identify the substances that are oxidized and reduced in the given reaction, let's analyze the changes in oxidation states:
The oxidation state of copper \((Cu)\) in \(CuO\) is \(+2\), and in \(Cu\) it is \(0\).
The oxidation state of hydrogen \((H)\) in \(H_2\) is \(0\), and in \(H_2O\) it is \(+1\).
In the reaction:
\[ \text{CuO} (s) + \text{H}_2 (g) \rightarrow \text{Cu} (s) + \text{H}_2 \text{O} (l) \]
\(CuO\) is reduced to \(Cu\).
\(H_2\) is oxidized to \(H_2O\).
So, the correct answer is: \(CuO\) is reduced and \(H_2\) gets oxidized.