indicate the oxidation state for each of the following:phosphorus in $\ce{p4o10}$sulfur in $\ce{h2so4}$nitrogen in $\ce{no2^-}$question 8in the reaction: $\ce{pbs + 3o2 - 2so2 + 2pbo}$which element is oxidized?which element is reduced?which substance is the oxidizing agent?which substance is the reducing agent?

Question

indicate the oxidation state for each of the following:phosphorus in $\ce{p4o10}$sulfur in $\ce{h2so4}$nitrogen in $\ce{no2^-}$question 8in the reaction: $\ce{pbs + 3o2 -> 2so2 + 2pbo}$which element is oxidized?which element is reduced?which substance is the oxidizing agent?which substance is the reducing agent?

Accepted Answer

# Explanation:
## Part 1: Oxidation state of phosphorus in $P_4O_{10}$
- **Step1: Set up the equation**
Let the oxidation state of $P$ be $x$. In $P_4O_{10}$, the oxidation state of $O$ is $- 2$. Using the rule that the sum of oxidation states in a neutral compound is $0$, we have $4x+10\times(-2)=0$.
- **Step2: Solve for $x$**
$$
\begin{align*}
4x - 20&=0\
4x&=20\
x&=+5
\end{align*}
$$
## Part 2: Oxidation state of sulfur in $H_2SO_4$
- **Step1: Set up the equation**
Let the oxidation state of $S$ be $y$. In $H_2SO_4$, the oxidation state of $H$ is $+1$ and of $O$ is $-2$. Using the rule that the sum of oxidation states in a neutral compound is $0$, we have $2\times(+1)+y + 4\times(-2)=0$.
- **Step2: Solve for $y$**
$$
\begin{align*}
2 + y-8&=0\
y - 6&=0\
y&=+6
\end{align*}
$$
## Part 3: Oxidation state of nitrogen in $NO_2^-$
- **Step1: Set up the equation**
Let the oxidation state of $N$ be $z$. In $NO_2^-$, the oxidation state of $O$ is $-2$. Using the rule that the sum of oxidation states in an ion is equal to the charge of the ion, we have $z+2\times(-2)=-1$.
- **Step2: Solve for $z$**
$$
\begin{align*}
z-4&=-1\
z&=+3
\end{align*}
$$
## Question 8:
### Oxidized element
- **Step1: Analyze oxidation states**
In $PbS$, $S$ has an oxidation state of $-2$. In $SO_2$, $S$ has an oxidation state of $+4$. Since the oxidation state of $S$ increases (from $-2$ to $+4$), $S$ is oxidized.
### Reduced element
- **Step1: Analyze oxidation states**
In $O_2$, $O$ has an oxidation state of $0$. In $SO_2$ and $PbO$, $O$ has an oxidation state of $-2$. Since the oxidation state of $O$ decreases (from $0$ to $-2$), $O$ is reduced.
### Oxidizing agent
- **Step1: Define oxidizing agent**
The oxidizing agent is the substance that causes oxidation (gets reduced). Since $O_2$ gets reduced (as shown above), $O_2$ is the oxidizing agent.
### Reducing agent
- **Step1: Define reducing agent**
The reducing agent is the substance that causes reduction (gets oxidized). Since $PbS$ gets oxidized (as $S$ in $PbS$ is oxidized), $PbS$ is the reducing agent.

# Answer:
- Phosphorus in $P_4O_{10}$: $+5$
- Sulfur in $H_2SO_4$: $+6$
- Nitrogen in $NO_2^-$: $+3$
- Oxidized element: $S$
- Reduced element: $O$
- Oxidizing agent: $O_2$
- Reducing agent: $PbS$

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QUESTION IMAGE

Question

Explanation:

Part 1: Oxidation state of phosphorus in \(P_4O_{10}\)

Part 2: Oxidation state of sulfur in \(H_2SO_4\)

Part 3: Oxidation state of nitrogen in \(NO_2^-\)

Question 8:

Oxidized element

Reduced element

Oxidizing agent

Reducing agent

Answer: