Oxidation: Loss of electrons, oxidation number increases. Reduction: Gain of electrons, oxidation number decreases.

Acidic: Balance using H⁺ ions. Basic: Balance as in acidic, then neutralize H⁺ with OH⁻ to form water.

Oxidation: Oxidation number increases. Reduction: Oxidation number decreases.

Oxidation: Electrons are lost. Reduction: Electrons are gained.

Oxidizing agent: Accepts electrons, gets reduced. Reducing agent: Donates electrons, gets oxidized.

1. Assign oxidation numbers. 2. Write half-reactions. 3. Balance elements (except O and H). 4. Balance O with H₂O. 5. Balance H with H⁺. 6. Balance charge with e⁻. 7. Combine half-reactions.

1. Balance as in acidic solution. 2. Neutralize H⁺ with OH⁻ to form H₂O. 3. Add the same amount of OH⁻ to the other side. 4. Simplify by canceling H₂O molecules.

1. Assign oxidation numbers. 2. Write half-reactions. 3. Balance elements (except O and H). 4. Balance oxygen atoms by adding H₂O molecules.

5. Balance hydrogen atoms by adding H⁺ ions. 6. Balance the charge by adding electrons. 7. Combine the half-reactions, canceling out any common species.

8. For basic solutions: Neutralize H⁺ ions with OH⁻ to form H₂O, and add the same number of OH⁻ to the other side. 9. Double-check that atoms and charges are balanced.

The atom undergoes oxidation, and its oxidation number increases.

The atom undergoes reduction, and its oxidation number decreases.

Water ($H_2O$) is formed, and you must add the same amount of $OH^−$ to the other side of the equation to maintain balance.

The overall equation will not be balanced, leading to incorrect stoichiometric coefficients and an inaccurate representation of the reaction.

$Cr_2O_7^{2−}$ is reduced to $Cr^{3+}$, and $I^−$ is oxidized to $I_2$.