When the following redox equation is balanced with smallest whole number coefficients, the coefficient for Sn(OH)₃⁻will be Bi(OH)₃(s) + Sn(OH)₃(aq) → Sn(OH)₆²⁻(aq) + Bi(s) (basic solution)

Question

Quizplus · Accepted Answer

First, we need to identify the oxidation states of each element in the equation: Bi: +3 Sn: +2 O: -2 H: +1 We then balance the equation by following these steps: 1. Balance the atoms that appear in both the half-reactions (in this case, Bi and Sn): Bi(OH)3 + Sn(OH)2 → Sn(OH)6 2- + Bi 2. Balance the number of electrons transferred in each half-reaction by multiplying one or both by coefficients: oxidation: Bi(OH)3 → Bi + 3H2O + 3e- reduction: Sn(OH)2 + 2e- → Sn(OH)6 2- 3. Combine the half-reactions by adding them together, canceling out electrons: Bi(OH)3 + Sn(OH)2 → Sn(OH)6 2- + Bi 2Bi(OH)3 + 3Sn(OH)2 → 3Sn(OH)6 2- + 2Bi At this point, all coefficients are integers, except for the coefficient of Sn(OH)2, which is 3. To make all coefficients integers, we can multiply each coefficient by 3, giving us: 6Bi(OH)3 + 9Sn(OH)2 → 9Sn(OH)6 2- + 6Bi Dividing all coefficients by 3, we end up with the balanced equation: 2Bi(OH)3 + 3Sn(OH)2 → 3Sn(OH)6 2- + 2Bi Therefore, the coefficient for Sn(OH)2 is 3, which is equivalent to a coefficient of 6 for Sn(OH)₃⁻. However, the question is asking for the coefficient of Sn(OH)₃, not Sn(OH)2, so the answer is 3.

When the Following Redox Equation Is Balanced with Smallest Whole