Assign oxidation states to each atom in each element, ion, or compound. Cl 2 , Fe 3+ , CuCl 2 , CH 4 , Cr 2 O 7 2– , HSO 4 –

Hey everyone welcome back. So let's get started with this video. So here they want us to see what is the oxidation state of each element in the following compound. Okay, so for number one, we have CH two, C. 02. So then the elements are see age and ceo we know the oxidation value for H is positive one, and C E O. Is negative one. Therefore, we can build an equation to solve for X, which is going to be representing carbon. So we're going to have X. Plus hydrogen is we have to And its oxidation value is positive one. Class C. L. We have to Observation Value is -1. The overall charge of this compound is zero. So then when we solve for X, we get zero, Southern Carbon is going to have an oxidation state of zero. Okay, so then let's move to number two. So here we have And each 4 - 04. So we're going to have to split this one into two. So we have NH four, which is the ammonium And we have CR 04, which is Chrome eight. The overall charge of ammonium Is a positive one. Don't get a charge of grooming is negative two. And that is why we need to so that we can have over a charge of zero. So elements here are going to be an aah cr and oh Once again, we know that hydrogen is going to have the value of plus one oxygen is going to have a value of negative two. So once again we can build equations So that we can solve for X. So let's go ahead and sulfur nitrogen and the NH four. So we have X. Which represents and plus H. There's four of them four times the charge of ages plus one. And the overall charge of ammonium is plus one. We solve for X. And we get -3. So then that is a charge of N. So then now we have to do is find the charge of C. R. And chromite. So then we have X. Which is going to represent C. R. Plus four, oxygen's we said is -2, And it's going to equal the overall charge of Chrome eight, which is -2. We solve for X. And we get positive six. So then we saw the chromium. Okay, so then now Let's move on to number three. So here we have carbonate and copper. Okay, so the overall charge of carbonate is negative two. Therefore, since this is a compound with the overall charge of zero, see you is going to have to be a positive too. So then see you is equal to positive too. Okay, so then the other elements we have is C. And oh, once again we said oxygen has a Oxidation state of -2. So then what we have to do is sulfur carbon. So then let's build an equation so that we can sulfur carbon. So then we have X. Which is going to represent carbon Plus three. There's three, oxygen's This charge is -2. The overall charge of carbonate is negative two. So then we solve for X. And we get positive for so then carbon is positive for Okay, so now we have B. A. Oh so be A. Oh So then oh we sent its oxidation state is -2. The overall charge of this compound is zero. So this is -2. This one is going to have to be positive too so that they can cancel out and have a charge of zero. Okay. And then finally we have K two cr 207. So the element is K. Cr and oh so we know the oxidation state of oxygen to be negative too. Case in group one A. So it's going to have a plus one so that we can build an equation to solve for C. R. So then here we have two crs therefore it's not only going to be X. It's going to be two X. Okay. And plus there's two. Okay to potassium. So too times plus one. And then here we have seven oxygen's That's seven times negative. two Overall charge of this compound is still zero. So then we bring everything down. We have two x. Plus two plus negative 14. Quarter zero, two Left 2 x. Is equal to positive. Well We divide by two. Leave that X. is equal to positive six and this is going to be for cr Okay, so we found the oxidation state of each element. Thank you for watching. I hope this helps, and I'll see you in the next video.

Ch.5 - Introduction to Solutions and Aqueous Solutions

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Ch.5 - Introduction to Solutions and Aqueous Solutions

Problem 62

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Chapter 5, Problem 62

Assign oxidation states to each atom in each element, ion, or compound. Cl₂, Fe³⁺, CuCl₂, CH₄, Cr₂O₇^2–, HSO₄^–

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For Cl2, since it is a diatomic molecule of the same element, each Cl atom has an oxidation state of 0.

For Fe3+, the oxidation state of Fe is +3, as indicated by the charge of the ion.

In CuCl2, assign the oxidation state of Cl as -1 (common for halogens), and use the fact that the compound is neutral to find the oxidation state of Cu.

For CH4, assign the oxidation state of H as +1 (common for hydrogen in compounds), and use the fact that the compound is neutral to find the oxidation state of C.

In Cr2O7^2-, assign the oxidation state of O as -2 (common for oxygen), and use the overall charge of the ion to find the oxidation state of Cr.

For HSO4-, assign the oxidation state of H as +1 and O as -2, then use the overall charge of the ion to find the oxidation state of S.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Oxidation States

Oxidation states, or oxidation numbers, are a way to keep track of electrons in chemical compounds. They indicate the degree of oxidation of an atom in a substance, helping to determine how electrons are transferred in redox reactions. The oxidation state of an atom can be positive, negative, or zero, depending on its electron gain or loss relative to its elemental form.

Rules for Assigning Oxidation States

There are specific rules for assigning oxidation states, such as: the oxidation state of an element in its standard state is zero, the oxidation state of a monoatomic ion equals its charge, and in compounds, hydrogen is usually +1 while oxygen is typically -2. These rules provide a systematic approach to determine the oxidation states of atoms in various compounds and ions.

Common Oxidation States of Elements

Different elements have common oxidation states that they typically exhibit in compounds. For example, alkali metals usually have an oxidation state of +1, alkaline earth metals +2, and transition metals can have multiple oxidation states. Understanding these common states aids in predicting the behavior of elements in chemical reactions and helps in assigning oxidation states accurately.