Answer:
b. 16, reactant side
Explanation:
Let's consider the following redox reaction.
MnO₄⁻(aq) + I⁻(aq) → Mn²⁺(aq) + I₂(s)
We can balance it using the ion-electron method.
Step 1: Identify both half-reactions
Reduction: MnO₄⁻(aq) → Mn²⁺(aq)
Oxidation: I⁻(aq) → I₂(s)
Step 2: Perform the mass balance, adding H⁺(aq) and H₂O(l) where appropriate
MnO₄⁻(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l)
2 I⁻(aq) → I₂(s)
Step 3: Perform the charge balance, adding electrons where appropriate
MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)
2 I⁻(aq) → I₂(s) + 2 e⁻
Step 4: Multiply both half-reactions by numbers so that the number of electrons gained and lost are equal
2 × (MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))
5 × (2 I⁻(aq) → I₂(s) + 2 e⁻)
Step 5: Add both half-reactions and cancel what is repeated on both sides
2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 e⁻ + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s) + 10 e⁻
The balanced reaction is:
2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)
