Question

1. complete and write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reactions below. if the substance is weak acid or base, use its molecular form when writing the equations.

chemical reactions	complete ionic equation	net ionic equation
b.) ch₃cooh(aq) + naoh (aq) →

1. give the oxidation number of the specified atom in the given molecules or ions.

molecules / ions	oxidation numbers	molecules / ions	oxidation numbers
b.) cl in hclo		e.) as in h₃aso₃
c.) br in bro₄⁻		f.) v in vo²⁺

1. for each of the following equations, identify the substance oxidized, the substance reduced, the oxidizing agent, and the reducing agent.

chemical reactions	substance oxidized	substance reduced	oxidizing agent	reducing agent
2.) 2 h₂s(g) + so₂(g) → 3 s(s) + 2 h₂o(l)

Explanation:

Step1: Write overall equations

Based on acid - base and neutralization reactions.

Step2: Write complete ionic equations

Dissociate strong electrolytes into ions.

Step3: Write net ionic equations

Cancel out spectator ions.

Step4: Calculate oxidation numbers

Use rules of oxidation number assignment.

Step5: Identify oxidized and reduced substances

Based on change in oxidation numbers.

Step6: Identify oxidizing and reducing agents

Oxidizing agent is the substance that is reduced, reducing agent is the substance that is oxidized.

Answer:

2.a)

• Overall Chemical Equation: $KOH(aq)+HCl(aq)

ightarrow KCl(aq) + H_2O(l)$

• Complete Ionic Equation: $K^+(aq)+OH^-(aq)+H^+(aq)+Cl^-(aq)

ightarrow K^+(aq)+Cl^-(aq)+H_2O(l)$

• Net Ionic Equation: $H^+(aq)+OH^-(aq)

ightarrow H_2O(l)$

2.b)

• Overall Chemical Equation: $CH_3COOH(aq)+NaOH(aq)

ightarrow CH_3COONa(aq)+H_2O(l)$

• Complete Ionic Equation: $CH_3COOH(aq)+Na^+(aq)+OH^-(aq)

ightarrow CH_3COO^-(aq)+Na^+(aq)+H_2O(l)$

• Net Ionic Equation: $CH_3COOH(aq)+OH^-(aq)

ightarrow CH_3COO^-(aq)+H_2O(l)$

3.a) Let the oxidation number of $Xe$ in $XeF_4$ be $x$. Fluorine has an oxidation - number of $- 1$. Then $x + 4\times(-1)=0$, so $x = + 4$.
3.b) Let the oxidation number of $Cl$ in $HClO$ be $x$. Hydrogen has an oxidation - number of $+1$ and oxygen has an oxidation - number of $-2$. Then $+1+x+( - 2)=0$, so $x = +1$.
3.c) Let the oxidation number of $Br$ in $BrO_4^-$ be $x$. Oxygen has an oxidation - number of $-2$. Then $x+4\times(-2)=-1$, so $x = +7$.
3.d) Let the oxidation number of $B$ in $B_2O_3$ be $x$. Oxygen has an oxidation - number of $-2$. Then $2x + 3\times(-2)=0$, so $x = +3$.
3.e) Let the oxidation number of $As$ in $H_3AsO_3$ be $x$. Hydrogen has an oxidation - number of $+1$ and oxygen has an oxidation - number of $-2$. Then $3\times(+1)+x+3\times(-2)=0$, so $x = +3$.
3.f) Let the oxidation number of $V$ in $VO^{2 + }$ be $x$. Oxygen has an oxidation - number of $-2$. Then $x+( - 2)=+2$, so $x = +4$.

4.1)

• Substance Oxidized: $NO$
• Substance Reduced: $O_3$
• Oxidizing Agent: $O_3$
• Reducing Agent: $NO$

4.2)

• Substance Oxidized: $H_2S$
• Substance Reduced: $SO_2$
• Oxidizing Agent: $SO_2$
• Reducing Agent: $H_2S$