Identify the oxidized reactant, the reduced reactant, the oxidizi...

Question

Identify the oxidized reactant, the reduced reactant, the oxidizing agent, and the reducing agent in the following reactions:
a. Fe(s)+Cu₂+(aq) → Fe₂+(aq)+Cu(s) 
b. Mg(s)+Cl₂(g) → MgCl₂(s)
c. 2 Al(s)+Cr₂O₃(s) → 2 Cr(s) + Al₂O₃(s)

Accepted Answer

Welcome back. Everyone on the reactant side in each of the given reactions below, determine the species that was oxidized and reduced and also provide the reducing agent and oxidizing agent. Let's begin with option one where we have on our reactant side, solid strontium reacting with solid iodine to produce strontium two, I died. So S R I sub two and this is a solid. We need to observe the oxidation states of each of our atoms beginning on our react inside the oxidation state of strontium in its standard state as a medal is going to be zero because recall that for the standard state of any atom, its oxidation state is always zero. Next for iodine, we observe that it's also in its elemental state as a di atomic solid. So it's going to have an oxidation state of zero being in its standard form. Now, on our product side, we have S R I S I sub to where the subscript of two belongs to the ion charge of strontium being our strontium two plus catalon. And then for iodine, we have the 1 -5 given by the subscript that is implied next to strontium of one. So recall that for the oxidation state of ions that would be equal to the ion charge. So for oxidation state of strontium on the product side, it's now equal to its ion charge of plus two. And for the oxidation state of I I died rather on the product side, it's equal to its anti on charge of minus one. So notice that strontium went from an oxidation state of zero to an oxidation state of plus two. So because we see that the oxidation state increased, this means that strontium lost electrons and therefore is oxidized, meaning that strontium is going to be the reducing agent. And now on the other hand, looking at I died, we observed that its oxidation state went from zero on the reactant side to -1 on the product side, meaning that its oxidation state decreased. So I died gained electrons and therefore is reduced. Meaning that idea is going to be our oxidizing agent. And so our final answers for part one is going to be that idea is reduced. And as the oxidizing agent and strontium was oxidized and is the reducing agent. So now moving on to part two, we have a different reaction where we have silver oxide A G sub 20 which is a solid reacting with solid sodium To form two moles of solid silver plus one mole of solid sodium oxide. So let's observe the oxidation states of our atoms. Beginning with silver, where on our reactant side, we have two moles of silver given by the subscript two Which have a oxidation state of its ion charge being plus one On oxygen. On the other hand, oxygen, we have one mole of and its ion charge is - given by that subscript of two next to silver. Because again, the ion charges crossed to form our silver oxide reactant and its oxidation state is equal to its anti on charge of minus two. Now, sodium on our reactant side has an oxidation state being in its standard state as a solid of zero. Now, on our product side, we have two moles of solid silver. So the oxidation state of silver in its elemental state or standard form is going to be zero on the product side. And now for sodium in our sodium oxide product, we have two moles of our a plus one catty on. And so the oxidation state is also equal to its ion charge of plus one. Whereas we have one mole of our oxygen or oh two minus oxide an ion and so its oxidation state is equal to its ion charge of minus two. So beginning with sodium on the react inside its oxidation state went from zero to on the product side plus one. So the oxidation state increased, meaning that sodium gained or rather sorry lost electrons, meaning that sodium is oxidized and is therefore going to be the reducing agent, silver. On the other hand, has an oxidation state of plus one on the reacting side and goes to an oxidation state of zero. On the product side, we see that the oxidation state decreased for silver. And so it gains electrons and therefore silver is reduced and is considered the oxidizing agent. So, we've identified our oxidizing agent and reducing agent for part two of our prompt. So it's highlighted in yellow are our next set of answers. And now we're going to move on to part three of our prompt where we consider the reaction where we have solid gallium. So, g a solid reacting with one mole of chromium three plus catalon to form our Galleon three plus cattle on the product side plus our chromium solid product. So let's observe the oxidation state of the atoms involved for gallium in its standard state as a solid on the react inside, it has an oxidation state equal to zero. Now, for chromium as a three plus carry on, it has an oxidation state equal to its caddy in charge of plus three. Now, on the product side, we have gallium as a three plus carry on. So its oxidation state is equal to its ion charge of plus three. And also on the product side, we have our solid chromium in its standard state as a metal. So it has an oxidation state of zero. So we can see that galleon went from having an oxidation state of zero on the reacting side to an oxidation state of plus three on the product side. And so the oxidation state is increased, meaning that gallium loses electrons and is therefore oxidized. And so our reducing agent is going to be gallium. Also observed that on our reacting side, we went from chromium with an oxidation state of plus three, two on the product side, chromium having an oxidation state of zero. So the oxidation state decreases, meaning that chromium gained electrons and is therefore reduced. And so our oxygen sorry oxidizing agent is going to be chromium in this case. And so our next set of answers are highlighted in yellow for the final part of our prompt. So all of our highlighted answers will correspond to choice d in the multiple choice as the final answer. I hope this made sense. And let us know if you have any questions.

Identify the oxidized reactant, the reduced reactant, the oxidizing agent, and the reducing agent in the following reactions: a. Fe(s)+Cu₂+(aq) → Fe₂+(aq)+Cu(s) b. Mg(s)+Cl₂(g) → MgCl₂(s) c. 2 Al(s)+Cr₂O₃(s) → 2 Cr(s) + Al₂O₃(s)

주요 개념

Oxidation and Reduction

Oxidizing and Reducing Agents

Half-Reactions

다음 시청