Question

4Au(s) + 16KCN(aq) + 3O2(g) + 6H2O(l) --> 4KAu(CN)4(aq) + 12KOH(aq) Balance using half reaction method. show all steps

Answer 1

O:

Au + KCN → KAu(CN)4

R:

O2 → 2H2O

b) Balance the oxygen atoms. Check if there are the same numbers of oxygen atoms on the left and right side, if they aren't equilibrate these atoms by adding water molecules.

O:

Au + KCN → KAu(CN)4

R:

O2 → 2H2O

c) Balance the hydrogen atoms. Check if there are the same numbers of hydrogen atoms on the left and right side, if they aren't equilibrate these atoms by adding protons (H+).

O:

Au + KCN → KAu(CN)4

R:

O2 + 4H+ → 2H2O

Step 4. Balance the charge. To balance the charge, add electrons (e-) to the more positive side to equal the less positive side of the half-reaction. It doesn't matter what the charge is as long as it is the same on both sides.

O: Au + KCN → KAu(CN)4

R: O2 + 4H+ + 4e- → 2H2O

Step 5. Make electron gain equivalent to electron lost. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. To make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.

O:4Au + 4KCN → 4KAu(CN)4

R: O2 + 4H+ + 4e- → 2H2O

Step 6. Add the half-reactions together. The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side.

4Au + O2 + 4KCN + 4H+ + 4e- → 4KAu(CN)4 + 2H2O

Step 7. Simplify the equation. The same species on opposite sides of the arrow can be canceled. Write the equation so that the coefficients are the smallest set of integers possible.

4Au + O2 + 4KCN + 4H+ + 4e- → 4KAu(CN)4 + 2H2O