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)₆^2-(aq) + Bi(s) (basic solution)

Question

Quizplus · Accepted Answer

The oxidation state of Bi changes from +3 to 0, and the oxidation state of Sn changes from +2 to +4. To balance the equation, we first balance the Bi by placing a coefficient of 2 in front of Bi(OH)3. This gives us:

2Bi(OH)3(s) + Sn(OH)2(aq) → Sn(OH)4(aq) + 2Bi(s)

Next, we balance the hydrogen by adding 6H+ to the right side:

2Bi(OH)3(s) + Sn(OH)2(aq) → Sn(OH)4(aq) + 2Bi(s) + 6H+

Finally, we balance the hydroxide ions by adding 3OH- to the left side:

2Bi(OH)3(s) + Sn(OH)2(aq) + 3OH-(aq) → Sn(OH)4(aq) + 2Bi(s) + 6H+

The coefficient for Sn(OH)2 is 1, which means the coefficient for Sn(OH)3- (which is formed by the reaction of Sn(OH)2 and OH-) is also 1. Therefore, the answer is C) 3.

When the Following Redox Equation Is Balanced with Smallest Whole