Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:

$N_{2} H_{4 (l)} + 2 H_{2} O_{2 (l)} \to N_{2 (g)} + 4 H_{2} O_{(l)}$

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Solution

The reaction is as follwos:

$N_{2} H_{4} (l) + 2 H_{2} O (l) \to N_{2} (g) + 4 H_{2} O (l)$

$N_{2} H_{4}$ is oxidized and $H_{2} O_{2}$ is reduced.

Hence, $H_{2} O_{2}$ is an oxidizing agent and $N_{2} H_{4}$ is a reducing agent.

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N_{2} H_{4} (l) + 2 H_{2} O_{2} (l) ⟶ N_{2} (g) + 4 H_{2} O (l)

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Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:N2H4(l)+2H2O2(l)→N2(g)+4H2O(l)

The reaction is as follwos:N2H4(l)+2H2O(l)→N2(g)+4H2O(l)N2H4 is oxidized and H2O2 is reduced.Hence, H2O2 is an oxidizing agent and N2H4 is a reducing agent.

Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:

$N_{2} H_{4 (l)} + 2 H_{2} O_{2 (l)} \to N_{2 (g)} + 4 H_{2} O_{(l)}$

The reaction is as follwos:

$N_{2} H_{4} (l) + 2 H_{2} O (l) \to N_{2} (g) + 4 H_{2} O (l)$

$N_{2} H_{4}$ is oxidized and $H_{2} O_{2}$ is reduced.

Hence, $H_{2} O_{2}$ is an oxidizing agent and $N_{2} H_{4}$ is a reducing agent.