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