Identify the substances that are oxidized and the substances that are reduced in the following reactions.
$(i) 4 Na (s) + O_{2} (g) \to 2 {Na}_{2} O (s) (ii) CuO (s) + H_{2} \to Cu (s) + H_{2} O (l)$

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Solution

$(i) 4 Na (s) + O_{2} (g) \to 2 {Na}_{2} O (s)$
In this reaction, Na is oxidized because it combines with $O_{2}$ to form ${Na}_{2} O . O_{2}$ is reduced because it is converted into ${Na}_{2} O$

. $(ii) CuO (s) + H_{2 (g)} \to Cu (s) + H_{2} O (l)$
In this reaction, CuO is reduced because it loses oxygen. $H_{2}$ is oxidized because it combines with oxygen of CuO to form water

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Similar questions

Identify the substances that are oxidised and the substances that are reduced in the following reactions.

(i)

(ii)

In the raction represented by the following equation :

$C u O (s) + H_{2}$ $⟶$ $C u (s) + H_{2} O (l)$

(a) name the substance oxidised (b) name the substance reduced

4 Na (s) + O_{2} (g) \to 2 {Na}_{2} O(s)

Identify the oxidised and reduced substances in the following reaction:

$4 N a (s) + O_{2} (g) \to 2 N a_{2} O (s)$

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