Balance and give electrons transferred.

Question

I consistently have gotten this question wrong.

When the following equation is balanced properly under acidic conditions, what are the coefficients of the species shown

__Al3+ + __SbO+ --> __Al + __Sb2O5

Water appears in the balanced equation as a __ (reactant, product, neither) with a coefficient of__ . (Enter 0 for neither.)

How many electrons are transferred in this reaction?__

Answer 1

Hi Chastity,

 

For completely balancing redox reactions, you should be following the following steps:

 

1. Split the reaction into oxidation and reduction half reactions.
1. In this case, the oxidation half is SbO →Sb₂O₅
2. The reduction half is Al³⁺→Al
2. Balance the atoms OTHER THAN H and O for each half
3. Balance the O by adding H₂O to the side that needs the O
4. Balance the H by adding H⁺ to the side that needs H
5. Balance the charges by adding electrons
1. The oxidation half reaction must have the electrons on the product side (right side of the arrow)
2. The reduction half reaction must have the electrons on the reactant side (left side of the arrow)
3. If either of these does not happen, there is a mistake! Examine the reactions, and correct any error.
6. Multiply each half reaction by the coefficient needed to get the number of electrons equal to the least common multiple of the electrons in the two half reactions.
7. Add the two half reactions together, and cancel items that are common to both sides.
8. DONE!

For this problem, the reduction half is trivial, being:

 

3e⁻ + Al³⁺ → Al

 

The oxidation half starts by balancing the antimony

 

2SbO → Sb₂O₅

 

followed by balancing the oxygen

 

3H₂O + 2SbO → Sb₂O₅

 

next the hydrogens

 

3H₂O + 2SbO → Sb₂O₅ + 6H⁺

 

finally the charges

 

3H₂O + 2SbO → Sb₂O₅ + 6H⁺ + 6e⁻

 

Steps 6 and 7 are what's left, and then you can answer the three parts of your problem.

 

Jim Scripko