ChemTopper Free Chemistry Help

Balance the following redox reaction in acidic solution:

CrO₄²⁻ + S²⁻ → Cr³⁺ + SO₄²⁻


1. Write the unbalanced equation in ionic form

CrO₄²⁻ + S²⁻ → Cr³⁺ + SO₄²⁻


2. Separate the equation into two half-reactions

Reduction half-reaction:

CrO₄²⁻ → Cr³⁺

Oxidation half-reaction:

S²⁻ → SO₄²⁻


3. Balance the atoms other than O and H

For the chromium half-reaction, Cr is already balanced:

CrO₄²⁻ → Cr³⁺

For the sulfur half-reaction, S is already balanced:

S²⁻ → SO₄²⁻


4. For reactions in acidic solution, add H₂O to balance O atoms and H⁺ to balance H atoms

Chromium half-reaction:

There are 4 O atoms on the left, so add 4H₂O to the right:

CrO₄²⁻ → Cr³⁺ + 4H₂O

There are now 8 H atoms on the right, so add 8H⁺ to the left:

8H⁺ + CrO₄²⁻ → Cr³⁺ + 4H₂O


Sulfur half-reaction:

There are 4 O atoms on the right, so add 4H₂O to the left:

S²⁻ + 4H₂O → SO₄²⁻

There are now 8 H atoms on the left, so add 8H⁺ to the right:

S²⁻ + 4H₂O → SO₄²⁻ + 8H⁺


5. Add electrons to one side of each half-reaction to balance the charges

Chromium half-reaction:

Charge on the left:

8(+1) + (−2) = +6

Charge on the right:

+3

Add 3e⁻ to the left:

3e⁻ + 8H⁺ + CrO₄²⁻ → Cr³⁺ + 4H₂O

This is reduction because electrons are gained.


Sulfur half-reaction:

Charge on the left:

−2

Charge on the right:

−2 + 8 = +6

Add 8e⁻ to the right:

S²⁻ + 4H₂O → SO₄²⁻ + 8H⁺ + 8e⁻

This is oxidation because electrons are lost.


6. Equalize the number of electrons in the two half-reactions

The two half-reactions contain 3e⁻ and 8e⁻.

LCM of 3 and 8 = 24

Multiply the chromium half-reaction by 8:

24e⁻ + 64H⁺ + 8CrO₄²⁻ → 8Cr³⁺ + 32H₂O

Multiply the sulfur half-reaction by 3:

3S²⁻ + 12H₂O → 3SO₄²⁻ + 24H⁺ + 24e⁻


7. Add the two half-reactions together and cancel common species

24e⁻ + 64H⁺ + 8CrO₄²⁻ → 8Cr³⁺ + 32H₂O

3S²⁻ + 12H₂O → 3SO₄²⁻ + 24H⁺ + 24e⁻

Cancel 24e⁻ from both sides.

Cancel 24H⁺ from both sides:

64H⁺ − 24H⁺ = 40H⁺

Cancel 12H₂O from both sides:

32H₂O − 12H₂O = 20H₂O

Therefore:

8CrO₄²⁻ + 3S²⁻ + 40H⁺ → 8Cr³⁺ + 3SO₄²⁻ + 20H₂O


8. Verify that the number of atoms and the charges are balanced

Atoms:

Cr: 8 = 8 ✓

S: 3 = 3 ✓

O:

Left = 8 × 4 = 32

Right = (3 × 4) + 20 = 32 ✓

H:

Left = 40

Right = 20 × 2 = 40 ✓


Charges on the left:

8(−2) + 3(−2) + 40(+1)

= −16 − 6 + 40

= +18


Charges on the right:

8(+3) + 3(−2)

= +24 − 6

= +18

Therefore, both atoms and charges are balanced.


FINAL BALANCED EQUATION

8CrO₄²⁻ + 3S²⁻ + 40H⁺ → 8Cr³⁺ + 3SO₄²⁻ + 20H₂O

Here is my question

balence the following half -reaction in acidic soltuion: Cro4-+s2-->Cr3+ + So42-

Both compounds are deprotonated at the sp³-hybridized α-carbon, not at oxygen or sulfur. The resulting conjugate base is stabilized by resonance, with the negative charge delocalized between the α-carbon and the heteroatom.

In compound I, the negative charge is delocalized onto oxygen, whereas in compound II, it is delocalized onto sulfur. Because oxygen is more electronegative than sulfur, it stabilizes the negative charge more effectively. Therefore, the conjugate base of compound I is more stable, making compound I more acidic than compound II.

Although sulfur is larger and more polarizable, that factor is more important when comparing negative charges located directly on oxygen and sulfur, such as RO⁻ versus RS⁻. Here, the strength of the carbonyl-like resonance stabilization makes the oxygen-containing compound more acidic.

So, the correct conclusion is:

Compound I is more acidic than compound II because its conjugate base is better stabilized by resonance involving oxygen.

For the attached question, I thought the most acidic molecule would be 2 because Sulfur can stabilize the electrons better than oxygen, since it is larger. However, the correct answer had the first molecule as the most acidic. I was wondering how that was the case.

Thank you so much Uma mam

Propane, C₃H₈, is a hydrocarbon. In its complete combustion, the second reactant is O₂, and the products are CO₂ and H₂O.
First, write and balance the equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Now check what is given and what is required.
We are given 1.00 g of propane, and we need to calculate the mass of O₂ consumed.
So, here is the plan:
grams of C₃H₈ → moles of C₃H₈ → moles of O₂ → grams of O₂
First, convert grams of propane to moles using its molar mass. Then convert moles of propane to moles of oxygen using the mole ratio from the balanced equation. Finally, convert moles of oxygen to grams using the molar mass of O₂.
The dimensional-analysis setup is:


All the intermediate units cancel, leaving the desired unit, grams of O₂.

3.63,g,O₂



Since 1.00 g has three significant figures, the final answer should also be reported to three significant figures.

I need help with the following:
**EP 3.14** Propane, C₃H₈, is a common fuel used for cooking and home heating. What mass of O₂ is consumed in the combustion of 1.00 g of propane?

a. 5.00 g

b. 0.726 g

c. 2.18 g

d. 3.63 g

Yes, you are right. The answer should be KNO₃. However, I am unable to see the full question, so I cannot confirm it with complete certainty.

Hi Uma mam, why isn't the answer KNO3 instead of NaCl? For NaCl, around 15 g is recovered and in KNO3, 105 g is recovered.