Treatment of gold metal with BrF₃ and KF produces Br₂ and KAuF₄, a salt of gold. Identify the oxidizing agent and the reducing agent in this reaction.

Video transcript

Hi everyone store asked to identify the reducing agent. An oxidizing agent. The reaction of gold medal with iodine In studying chloride producing study in touch with your right and sodium iodide, we're gonna have a U. For gold plus I two plus N A C L. And it's gonna yield in a AUCL. four plus in A. I. And recall that the loss of electrons gonna be oxidation which is going to be our reducing agent in the game of electrons. It's gonna be reduction. Which is gonna be oxidizing agent. And for the changes in oxidation states, we're gonna look at gold Over here, it's zero And then over here clearly is gonna have a negative one. Siding is going to have a plus one. So we got negative four Plus one over here. This has to be a Plus three. So gold has lost electrons. It's going to be a reducing agent. And for iodine Over zero over here I have a plus one over here and a -1 over here. So this gained electrons, it's gonna be an oxidizing agent. And for corinne We're gonna have a negative one here. Plus one here. So it's gonna be negative one And the -1. So there's no change in facility in. We have a plus one here and plus one over here as well. So this is no change to our reducing agent, it's going to be gold. And our oxidizing agent is iodine. Thanks for watching my video and I hope it was helpful