Answer: 5
Step-by-step explanation:
To balance the oxidation-reduction reaction in a basic solution, you need to follow these steps:
Step 1: Separate the reaction into two half-reactions, one for oxidation and one for reduction.
H₂S (aq) → S(s) (oxidation)
Cr₂O₇²⁻ (aq) → Cr³⁺ (aq) (reduction)
Step 2: Balance the atoms in each half-reaction, excluding oxygen and hydrogen.
For the oxidation half-reaction:
2H₂S (aq) → S(s)
For the reduction half-reaction:
Cr₂O₇²⁻ (aq) → 2Cr³⁺ (aq)
Step 3: Balance the oxygen atoms by adding water (H₂O) to the side that needs it.
For the oxidation half-reaction, we don't need to balance oxygen.
For the reduction half-reaction, we need to add 7H₂O to the left side:
Cr₂O₇²⁻ (aq) + 14H⁺(aq) → 2Cr³⁺ (aq) + 7H₂O (l)
Step 4: Balance the hydrogen atoms by adding H⁺ ions to the side that needs it.
For the oxidation half-reaction, we need to add 4H⁺ to the left side:
4H⁺(aq) + 2H₂S (aq) → S(s)
For the reduction half-reaction, we already balanced the hydrogen atoms in step 3.
Step 5: Balance the charges by adding electrons (e⁻) to the side that needs it.
For the oxidation half-reaction, we need to add 8e⁻ to the left side:
8e⁻ + 4H⁺(aq) + 2H₂S (aq) → S(s)
For the reduction half-reaction, we need to add 6e⁻ to the left side:
Cr₂O₇²⁻ (aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺ (aq) + 7H₂O (l)
Step 6: Equalize the number of electrons in both half-reactions by multiplying them as necessary.
We need to multiply the oxidation half-reaction by 3 to equalize the electrons:
24e⁻ + 12H⁺(aq) + 6H₂S (aq) → 3S(s)
Step 7: Combine the half-reactions and cancel out any common terms.
24e⁻ + 12H⁺(aq) + 6H₂S (aq) + Cr₂O₇²⁻ (aq) + 14H⁺(aq) + 6e⁻ → 3S(s) + 2Cr³⁺ (aq) + 7H₂O (l)
Simplifying and canceling terms, we get:
18e⁻ + 20H⁺(aq) + 6H₂S (aq) + Cr₂O₇²⁻ (aq) → 3S(s) + 2Cr³⁺ (aq) + 7H₂O (l)
Finally, the sum of the coefficients of the products is 3 + 2 = 5. Therefore, the correct answer is 5.