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

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