Balance the following redox reations:
K2Cr2O7 + Nal + H2SO4 --- Cr2(SO4)3 + I2 + H2O + Na2SO4 + K2SO4
See above.
To balance a redox reaction, you need to ensure that the number of atoms and charges are balanced on both sides of the equation. Here's how you can balance this given redox reaction step-by-step:
1. Split the reaction into two half-reactions: one for the oxidation (loss of electrons) and one for the reduction (gain of electrons).
Half-reaction 1 (Oxidation):
K2Cr2O7 → Cr2(SO4)3
Half-reaction 2 (Reduction):
Nal + H2SO4 → I2 + H2O + Na2SO4 + K2SO4
2. Balance the atoms in each half-reaction, except for oxygen and hydrogen.
Half-reaction 1 (Oxidation):
K2Cr2O7 → Cr2(SO4)3
Half-reaction 2 (Reduction):
2Nal + H2SO4 → I2 + H2O + Na2SO4 + K2SO4
3. Balance the number of oxygen atoms by adding water (H2O).
Half-reaction 1 (Oxidation):
K2Cr2O7 → Cr2(SO4)3
Half-reaction 2 (Reduction):
2Nal + H2SO4 → I2 + H2O + Na2SO4 + K2SO4
4. Balance the number of hydrogen atoms by adding hydrogen ions (H+).
Half-reaction 1 (Oxidation):
K2Cr2O7 → Cr2(SO4)3
Half-reaction 2 (Reduction):
2Nal + H2SO4 → I2 + 2H2O + Na2SO4 + K2SO4
5. Balance the charge in each half-reaction by adding electrons (e-).
Half-reaction 1 (Oxidation):
K2Cr2O7 → Cr2(SO4)3
Half-reaction 2 (Reduction):
2Nal + 2H+ + H2SO4 → I2 + 2H2O + Na2SO4 + K2SO4 + 2e-
6. Equalize the number of electrons in both half-reactions by multiplying each half-reaction by an appropriate number, ensuring that the total number of electrons gained in the reduction half-reaction equals the total number of electrons lost in the oxidation half-reaction. In this case, multiply the reduction half-reaction by 6 and the oxidation half-reaction by 3.
Half-reaction 1 (Oxidation):
3K2Cr2O7 → 3Cr2(SO4)3
Half-reaction 2 (Reduction):
12Nal + 12H+ + 6H2SO4 → 6I2 + 12H2O + 6Na2SO4 + 6K2SO4 + 12e-
7. Combine the two half-reactions by canceling out electrons on both sides.
3K2Cr2O7 + 12Nal + 12H+ + 6H2SO4 → 3Cr2(SO4)3 + 6I2 + 12H2O + 6Na2SO4 + 6K2SO4
8. Simplify the equation by removing any common factors and ensuring the smallest whole-number coefficients.
K2Cr2O7 + 6Nal + 6H+ + 3H2SO4 → 3Cr2(SO4)3 + 3I2 + 6H2O + 3Na2SO4 + 3K2SO4
Finally, the balanced equation is:
K2Cr2O7 + 6Nal + 6H+ + 3H2SO4 → 3Cr2(SO4)3 + 3I2 + 6H2O + 3Na2SO4 + 3K2SO4