In the reaction represented by the equation CuO + H₂Cu + H₂O

a) Name the substance oxidized and the substance reduced

b) Name the oxidizing agent and the reducing agent

Open in App

Solution

CuO + H₂ ---------> Cu + H₂O

Here Cu is gaining electrons and hence is going from Cu²⁺ to Cu hence is getting reduced.

where as H is going from 0 to -1 hence oxidised.

the species getting reduced is called as oxidising agent i.e Cu
and H is the reducing agent.

Suggest Corrections

Similar questions

CuO (s) + H_{2} (g) \underset{}{\to} Cu (s) + H_{2} O (1)

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

3 M n O_{2} + 4 A l \to 3 M n + 2 A l_{2} O_{3}

Identify the substance oxidized, substance reduced, oxidizing agent and reducing agent in the given reaction
$H_{2} S + B r_{2} \to 2 H B r + S$

In the reaction, H₂S + Cl₂ → 2HCl + S; hydrogen sulphide is:

Join BYJU'S Learning Program