Balancing Redox Reactions in Acidic Solution, Answer Nos. 11-15

11)

the final answer:

12)

multiply top half-reaction by 3, bottom by 2; the final answer:

13) This problem poses interesting problems, especially with the Cl. The key to solving ths problem is to eliminate everything not directly involved in the redox. That means the H in HFeCl₄ as well as the Cl in it and HCl. When we do that, this is the unbalanced, ionic form we wind up with:

the half-reactions (already balanced) are as follows:

the final answer:

We will go back to the molecular equation with 8HCl. Six of the HCl molecules supply the 6H⁺ going to 3H₂. The 7th and 8th HCl molecules supply the two H present in 2HFeCl₄, as well as the 8 Cl needed in the two HFeCl₄ molecules.

14) the half-reactions (already balanced) are as follows:

multiply the top half-reaction by 2 and the bottom one by 3, add them and eliminate 6H⁺:

you can combine the hydrogen ion and the nitrate ion like this:

This creates a molecular equation, but ignores the reality that nitric acid is a strong acid and is fully ionized in solution.

15) these are the half-reactions:

Only the second half-reaction needs to be multiplied through by a factor, then we add the two half-reactions for the final answer: