Identify the oxidizing agent and the reducing agent in the reaction.

Question

Identify the oxidizing agent and the reducing agent in the reaction.

8H⁺(aq) + 6Cl^–(aq) + Sn(s) + 4NO₃^–(aq) → SnCl₆^2–(aq) + 4NO₂(g) + 4H₂O(l)

the answer is:

oxidizing agent: NO₃^-

reducing agent: Sn

please show steps and explain so I can follow, thank you!

Answer 1

8H⁺(aq) + 6Cl^-(aq) + Sn(s) + 4NO₃^-(aq) → SnCl₆^2-(aq) + 4NO₂(aq) + 4H₂O(l)

+1 -1 0 +5-2 +4-1 +4-2 +1-2 oxidation states

To find the oxidation state of SnCl₆^2- set Sn equal to x and Cl equal to -6 and whole equation equal to -2 and solve for x:

x-6=-2

x=+4

Do the same for NO₃^- and NO₂. H and Cl states are due to position on periodic table. Sn(s) is a solid. Water is neutral molecule.

Sn lost 4 e^- making it the reducing agent because it got oxixized.

Change in oxidation state for NO₃^- as it went to NO₂ makes it the oxidizing agent meaning it got reduced.

Answer 2

8H⁺(aq) + 6Cl^–(aq) + Sn(s) + 4NO₃^–(aq) ==> SnCl₆^2–(aq) + 4NO₂(g) + 4H₂O(l)

Look at the oxidation number of each element on the left and right side.

H is 1+ on both sides

Cl is 1- on both sides

Sn is zero on the left side and 4+ on the right side. Sn has been oxidized so it is the reducing agent

N in NO₃^- is 5+ on the left side is 4+ on the right side in NO₂. N in NO₃^- has been reduced so is the oxidizing agent.