Question

1. determine the oxidation numbers for each element.

a. br₂
br = 0
d. so₃²⁻
s = + 4
o₃²⁻=-2

1. determine the substance being oxidized, the oxidizing agent in the following reactions.

a. 2k(s)+br₂(l)→2kbr(s)
b. ca(s)+2h⁺(aq)→ca²⁺(aq)+h₂(g)
c. fe₂o₃(s)+3co(g)→2fe(s)+3co₂(g)

Explanation:

Step1: Recall oxidation - number rules

In a free element, the oxidation number of each atom is 0. For mono - atomic ions, the oxidation number is equal to the charge of the ion. In compounds, the sum of oxidation numbers of all atoms equals the charge of the species.

Step2: Analyze reaction a: $2K(s)+Br_2(l)\to2KBr(s)$

• Oxidation number of $K$ in $K(s)$ is 0. In $KBr$, $K$ has an oxidation number of + 1 (since it is a Group 1 metal). So $K$ is oxidized (its oxidation number increases from 0 to + 1).
• Oxidation number of $Br$ in $Br_2$ is 0. In $KBr$, $Br$ has an oxidation number of - 1. So $Br_2$ is the oxidizing agent (it causes the oxidation of $K$ and is itself reduced).

Step3: Analyze reaction b: $Ca(s)+2H^+(aq)\to Ca^{2 + }(aq)+H_2(g)$

• Oxidation number of $Ca$ in $Ca(s)$ is 0. In $Ca^{2+}(aq)$, the oxidation number of $Ca$ is + 2. So $Ca$ is oxidized.
• Oxidation number of $H$ in $H^+$ is + 1. In $H_2$, the oxidation number of $H$ is 0. So $H^+$ is the oxidizing agent.

Step4: Analyze reaction c: $Fe_2O_3(s)+3CO(g)\to2Fe(s)+3CO_2(g)$

• In $Fe_2O_3$, let the oxidation number of $Fe$ be $x$. Since $O$ has an oxidation number of - 2, we have $2x+3\times(-2)=0$, solving for $x$ gives $x = + 3$. In $Fe(s)$, the oxidation number of $Fe$ is 0. So $Fe_2O_3$ is reduced and is the oxidizing agent.
• In $CO$, let the oxidation number of $C$ be $y$. Since $O$ has an oxidation number of - 2, we have $y+( - 2)=0$, so $y = + 2$. In $CO_2$, the oxidation number of $C$ is + 4. So $CO$ is oxidized.

Answer:

a. Substance being oxidized: $K$, Oxidizing agent: $Br_2$
b. Substance being oxidized: $Ca$, Oxidizing agent: $H^+$
c. Substance being oxidized: $CO$, Oxidizing agent: $Fe_2O_3$