Question

study check
identify the oxidizing agent and reducing agent in each of the following:
a) 2h₂(g) + o₂(g) → 2h₂o (g)
b) cu(s) + 4hno₃(aq) → cu(no₃)₂(aq) + 2no₂(g) + 2h₂o (l)

Explanation:

Step1: Analyze oxidation states in reaction a

• In \(H_2\), \(H\) has an oxidation state of \(0\). In \(H_2O\), \(H\) has an oxidation state of \(+ 1\). So \(H_2\) is oxidized (loses electrons).
• In \(O_2\), \(O\) has an oxidation state of \(0\). In \(H_2O\), \(O\) has an oxidation state of \(-2\). So \(O_2\) is reduced (gains electrons).

Step2: Analyze oxidation states in reaction b

• In \(Cu\), \(Cu\) has an oxidation state of \(0\). In \(Cu(NO_3)_2\), \(Cu\) has an oxidation state of \(+2\). So \(Cu\) is oxidized (loses electrons).
• In \(HNO_3\), \(N\) has an oxidation state of \(+5\). In \(NO_2\), \(N\) has an oxidation state of \(+4\). So \(HNO_3\) is reduced (gains electrons).

Answer:

a) Oxidizing agent: \(O_2\); Reducing agent: \(H_2\)
b) Oxidizing agent: \(HNO_3\); Reducing agent: \(Cu\)