Math / Algebra

Question3. Which of the following is NOT a redox reaction? A. $\mathrm{CH}_{1}(8)+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{HCl}(g)$ B. $\mathrm{C}(\mathrm{S})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})$ $\mathrm{C}_{2} \mathrm{CO}(g) \rightarrow \mathrm{CO}_{2}(g)+C(s)$ D. $\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{NaOH}(a q) \rightarrow \mathrm{CH}_{3} \mathrm{COONa}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}$ (l)
4. What are the axidation numbers of the elements in sulfuric acid, $\mathrm{H}_{2} \mathrm{SO}_{4}$ ?

Studdy Solution

STEP 1

What is this asking? We need to find which reaction isn't a redox reaction (where oxidation states change) and figure out the oxidation numbers of each element in sulfuric acid. Watch out! Don't forget, even though elements in their standard states have an oxidation number of zero, they can have different oxidation numbers in compounds!

STEP 2

1. Analyze the reactions
2. Determine oxidation numbers in sulfuric acid

STEP 3

Let's start with reaction A: $\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{CH}_{3}\mathrm{Cl}(g)+\mathrm{HCl}(g)$ .
In $\mathrm{CH}_{4}$ , carbon has an oxidation number of -4 and hydrogen has +1.
In $\mathrm{Cl}_{2}$ , chlorine is 0.
In $\mathrm{CH}_{3}\mathrm{Cl}$ , carbon is -2, hydrogen is +1, and chlorine is -1.
In $\mathrm{HCl}$ , hydrogen is +1 and chlorine is -1.
See how the oxidation states of carbon and chlorine changed?
That means it's a redox reaction!

STEP 4

Next up, reaction B: $\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)$ .
Carbon goes from 0 to +4 and oxygen goes from 0 to -2.
Oxidation states changed, so this is also redox!

STEP 5

Now for reaction C: $\mathrm{CO}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{C}(s)$ .
In $\mathrm{CO}$ , carbon is +2 and oxygen is -2.
In $\mathrm{CO}_{2}$ , carbon is +4 and oxygen is -2.
And the elemental carbon is 0.
Carbon changes from +2 to both +4 and 0.
Redox alert!

STEP 6

Finally, reaction D: $\mathrm{CH}_{3}\mathrm{COOH}(aq)+\mathrm{NaOH}(aq) \rightarrow \mathrm{CH}_{3}\mathrm{COONa}(aq)+\mathrm{H}_{2}\mathrm{O}(l)$ .
This is an acid-base reaction.
No oxidation state changes occur here.
This is NOT a redox reaction.

STEP 7

We want to find the oxidation numbers in $\mathrm{H}_{2}\mathrm{SO}_{4}$ .
Remember, hydrogen is usually +1 and oxygen is usually -2.

STEP 8

Let's set up an equation.
We have 2 hydrogens, so that's $2 \cdot (+1)$ .
We also have 4 oxygens, so that's $4 \cdot (-2)$ .
Let's call the oxidation number of sulfur $x$ .
The overall charge of the molecule is 0, so we have: $2 \cdot (+1) + x + 4 \cdot (-2) = 0$ .

STEP 9

Simplifying, we get $2 + x - 8 = 0$ , which means $x - 6 = 0$ .
Solving for $x$ , we find $x = +6$ .
So, sulfur's oxidation number is +6!

STEP 10

For question 3, the reaction that is NOT a redox reaction is D.
For question 4, the oxidation numbers are: H = +1, S = +6, and O = -2.

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Studdy Solution

STEP 1

STEP 2

1. Analyze the reactions
2. Determine oxidation numbers in sulfuric acid

STEP 3

STEP 4

Next up, reaction B: $\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)$ .
Carbon goes from 0 to +4 and oxygen goes from 0 to -2.
Oxidation states changed, so this is also redox!

STEP 5

Now for reaction C: $\mathrm{CO}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{C}(s)$ .
In $\mathrm{CO}$ , carbon is +2 and oxygen is -2.
In $\mathrm{CO}_{2}$ , carbon is +4 and oxygen is -2.
And the elemental carbon is 0.
Carbon changes from +2 to both +4 and 0.
Redox alert!

STEP 6

Finally, reaction D: $\mathrm{CH}_{3}\mathrm{COOH}(aq)+\mathrm{NaOH}(aq) \rightarrow \mathrm{CH}_{3}\mathrm{COONa}(aq)+\mathrm{H}_{2}\mathrm{O}(l)$ .
This is an acid-base reaction.
No oxidation state changes occur here.
This is NOT a redox reaction.

STEP 7

We want to find the oxidation numbers in $\mathrm{H}_{2}\mathrm{SO}_{4}$ .
Remember, hydrogen is usually +1 and oxygen is usually -2.

STEP 8

Let's set up an equation.
We have 2 hydrogens, so that's $2 \cdot (+1)$ .
We also have 4 oxygens, so that's $4 \cdot (-2)$ .
Let's call the oxidation number of sulfur $x$ .
The overall charge of the molecule is 0, so we have: $2 \cdot (+1) + x + 4 \cdot (-2) = 0$ .

STEP 9

Simplifying, we get $2 + x - 8 = 0$ , which means $x - 6 = 0$ .
Solving for $x$ , we find $x = +6$ .
So, sulfur's oxidation number is +6!

STEP 10

For question 3, the reaction that is NOT a redox reaction is D.
For question 4, the oxidation numbers are: H = +1, S = +6, and O = -2.

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Studdy Solution

STEP 1

STEP 2

1. Analyze the reactions2. Determine oxidation numbers in sulfuric acid

STEP 3

STEP 4

Next up, reaction B: C(s)+O2(g)→CO2(g)\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)C(s)+O2​(g)→CO2​(g).Carbon goes from **0** to **+4** and oxygen goes from **0** to **-2**.Oxidation states changed, so this is also redox!

STEP 5

STEP 6

STEP 7

We want to find the oxidation numbers in H2SO4\mathrm{H}_{2}\mathrm{SO}_{4}H2​SO4​.Remember, hydrogen is usually **+1** and oxygen is usually **-2**.

STEP 8

STEP 9

Simplifying, we get 2+x−8=02 + x - 8 = 02+x−8=0, which means x−6=0x - 6 = 0x−6=0.Solving for xxx, we find x=+6x = +6x=+6.So, sulfur's oxidation number is **+6**!

STEP 10

For question 3, the reaction that is NOT a redox reaction is **D**.For question 4, the oxidation numbers are: H = **+1**, S = **+6**, and O = **-2**.

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1. Analyze the reactions
2. Determine oxidation numbers in sulfuric acid

Next up, reaction B: $\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)$ .
Carbon goes from 0 to +4 and oxygen goes from 0 to -2.
Oxidation states changed, so this is also redox!

We want to find the oxidation numbers in $\mathrm{H}_{2}\mathrm{SO}_{4}$ .
Remember, hydrogen is usually +1 and oxygen is usually -2.

Simplifying, we get $2 + x - 8 = 0$ , which means $x - 6 = 0$ .
Solving for $x$ , we find $x = +6$ .
So, sulfur's oxidation number is +6!

For question 3, the reaction that is NOT a redox reaction is D.
For question 4, the oxidation numbers are: H = +1, S = +6, and O = -2.