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Identify the substances that are oxidised and the substances that are reduced in the following reaction: CuO(s) + H2(g) → Cu(s) + H2O(l) |
CuO is reduced and H2 gets oxidised CuO is oxidised and H2 gets reduced Both CuO as well as H2 is oxidised Both CuO as well as H2 is reduced |
CuO is reduced and H2 gets oxidised |
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. |
