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An equilibrium is established according to the following equation$\mathrm{Hg}_{2}^{2+}(a q)+\mathrm{NO}_{3}^{-}(a q)+3 \mathrm{H}^{+}(a q) \rightleftharpoons 2 \mathrm{Hg}^{2+}(a q)+\mathrm{HNO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \quad K_{c}=4.6$What will happen in a solution that is 0.20$M$ each in $\mathrm{Hg}_{2}^{2+}, \quad \mathrm{NO}_{3}^{-}, \quad \mathrm{H}^{+}, \mathrm{Hg}^{2+},$ and $\mathrm{HNO}_{2} ?$(a) $\mathrm{Hg}_{2}^{2+}$ will be oxidized and $\mathrm{NO}_{3}^{-}$ reduced.(b) $\mathrm{Hg}_{2}^{2+}$ will be reduced and $\mathrm{NO}_{3}$ - oxidized.(c) $\mathrm{Hg}^{2+}$ will be oxidized and $\mathrm{HNO}_{2}$ reduced.(d) $\mathrm{Hg}^{2+}$ will be reduced and $\mathrm{HNO}_{2}$ oxidized.(e) There will be no change because all reactants and products have an activity of $1 .$

a) this option is wrong.b) this option is wrong.c) this option is wrong.d) this option is correct.e) this option is wrong.

Chemistry 102

Chapter 13

Fundamental Equilibrium Concepts

Chemical Equilibrium

Aqueous Equilibria

University of Central Florida

University of Kentucky

University of Toronto

Lectures

00:41

In chemistry, an ion is an atom or molecule that has a non-zero net electric charge. The name was coined by John Dalton for ions in 1808, and later expanded to include molecules in 1834.

10:03

In thermodynamics, a state of thermodynamic equilibrium is a state in which a system is in thermal equilibrium with its surroundings. A system in thermodynamic equilibrium is in thermal equilibrium, mechanical equilibrium, electrical equilibrium, and chemical equilibrium. A system is in equilibrium when it is in thermal equilibrium with its surroundings.

03:37

Given that\[\begin…

05:37

The equilibrium constant f…

03:30

Given the following reduct…

11:00

06:57

A disproportionation react…

07:00

Calculate AG? ad the equil…

this question says an equilibrium is established according to the following reaction, which I've pasted in here. That, and asked what will happen in a solution that is point to Moller in each of the ions. So each of the A clear solutions is point to Mueller. So what we need to do to solve this problem is we need to figure out the reaction quotient to see where we are in the reaction. So this is Q. See the reaction quotient, and depending on whether or not QC is greater than or less than Casey, we can figure out where they're not. The ration goes left or right, and then what's being oxidized on what's being reduced? So this is the equilibrium question. It's the concentration of the products H G two plus squared is there's a to here times a concentration of h N 02 divided by the concentration of H G to two plus times concentration of enter three minus times, concentration of age plus cubed because there's this three here and itself tell us that we're 30.2 in all of these ions, so we can simplify this by saying this is basically the same thing as saying This is point. Um, Was that point, too? This would be cubed since there three of them on top. Why? To buy 0.2. And this is one too plus three. Morse five. So I went to to the fifth. You're simplifying that a little bit. You saw this. You would get Q C equal to 25. Therefore, Q is greater than Casey. And when this is the case, we're going to shift the reaction to the left towards the reactant CE, so we never go into the left. But all of the multiple choice questions here tell us something is being oxidized or reduced so real quickly. I'm gonna write the oxidation states of the major ions. If H G two is two plus each of those H G's has to be plus one, oxygen's always minus two. There are three of them's that would give us minus six. Therefore, the nitrogen must be plus five. These hydrogen, the course air plus one. Here we have HD two plus. Those will both be that that is now, since there's only one of them that'll be plus two oxidation. State oxygen is minus two. There are two of them that's minus four. Hydrogen gets us plus one. Since this is neutral, we need the other three while the charges to come from the nitrogen summations plus three and murder ignore water. Since that's not an eye on here. Okay, so now we know what the oxidation states are. Let's see what's being oxides and what's being reduced. So we know we're going towards the left or shifting the reaction. Manship backed the black pen Here, we're going this way. So how would we describe this? Here we have hte G two plus going from a plus to oxidation State to a minus two a plus one oxidation state. So it's gaining an electron. It is reduced. So I'm gonna write that H G two plus is reduced. Okay. And let's see here the nitrogen in h N 02 goes from plus three two plus five. It's losing electrons. So then we would also say that each and No. Two is oxidized. So that's how I would describe what's gonna happen when we have those concentrations of ions shifting to the left. HD two plus will be reduced. A Jenna to will be oxidized, and this is described perfectly by option choice D

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