Balance the following half-reactions. (b) (basic) CrO 4 2- (aq) → Cr(OH) 4 - (aq)

Welcome back, everyone. We need to consider the following half reaction where lead to oxide solid forms lead two plus catalon and need to balance this half reaction in basic solution. So, recall that we would begin by balancing non hydrogen and non oxygen atoms first. And so we want to balance out our lead atoms. Now note that on both sides of the equation, we have one mole of lead. So now we would move on to balance oxygen. So let's rewrite our reaction just so everything is a bit bigger. So balancing out oxygen. Next, we see that we have two moles of oxygen on the reactant side and none on the product side, recall that we balance oxygen using water. And so we would go ahead and add two moles of water to the product side so that we now have oxygen balanced. But recognized that with the addition of water on the product side, we've now introduced a total of four moles of hydrogen on the product side. And so recall that we can balance hydrogen using protons. And so we would expand our react inside and add a total of four moles of protons to the products or to the reactant side. Now, with all of our Adams balanced, we want to make sure we have net charge balanced. And so looking at the reacting side, we have a net charge of plus four. And on the product side, we have a net charge of plus two. So in order to balance out our charges, we add electrons to the more positive side, which would be our reactant side here. And so we would add a total of two electrons so that we have a net charge of plus two on both sides of our reaction, meaning our atoms and ions, sorry, our atoms and net charge is balanced for our entire reaction. Now with Adams and charge balanced, we want to recognize that we're balancing in basic solution and because we added protons to our react inside and sorry, just to make sure our work is visible, let's resize this. So going back to what I was saying, because we added four moles of protons to the react inside. And because we're bouncing in basic solution, we add hydroxide to the side or we add hydroxide to both sides of our equation. And so we're going to expand our react inside again And we're going to need to add an equivalent amount of hydroxide. So plus four moles and let's actually let's move this over a bit more So that we can add four moles of hydroxide. and because we added four moles of hydroxide to the react inside. We must add the same amount to the product side. So let's move our notes up And we will add four moles of hydroxide to the product side. Now recall that when we have a plus and O H, they're going to combine to form water and so on are reacting side, we can now rewrite it so that we have our lead to oxide plus for moles of protons with four multi hydroxide will form for moles of water Plus our two electrons Which yields lead two plus Two moles of water on the product side and four moles of hydroxide on the product side still. And now we want to cancel out what we have in common on both sides of the reaction. And so recognize that we have two moles of water on the product side in which we can count two out of the four moles of water on the react inside leaving us with two moles of water on the reactant side. Now, now nothing else is common. And so we can write our overall reaction in which we have one mole of lead to oxide solid plus two moles of liquid water plus two electrons yields one mole of lead, two plus cat ion plus four moles of hydroxide. And this entire equation is going to be our final answer as our overall reaction which is balanced in basic solution. I hope that we're sorry. Our half reaction balanced in basic solution. So I hope that everything I went through is clear. If you have any questions, please leave them down below. Our final answer corresponds to choice D and I'll see everyone in the next practice video.

Ch.19 - Electrochemistry

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McMurry 8th Edition

Ch.19 - Electrochemistry

Problem 48b

Chapter 19, Problem 48b

Balance the following half-reactions. (b) (basic) CrO₄^2-(aq) → Cr(OH)₄^-(aq)

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Identify the elements that are being oxidized and reduced. In this case, chromium (Cr) is the element of interest.

Balance the chromium atoms on both sides of the equation. Since there is one Cr atom on each side, they are already balanced.

Balance the oxygen atoms by adding water (H2O) molecules. There are 4 oxygen atoms on the left and 4 on the right, so they are balanced.

Balance the hydrogen atoms by adding hydroxide ions (OH^-) to the side that needs more hydrogen. Add 4 OH^- to the left side to balance the 4 H in Cr(OH)4^- on the right.

Balance the charge by adding electrons (e^-). The left side has a charge of -6 (from CrO4^2- and 4 OH^-), and the right side has a charge of -1 (from Cr(OH)4^-), so add 3 e^- to the left side to balance the charges.

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Key Concepts

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

Half-Reaction Method

The half-reaction method is a technique used to balance redox reactions by separating the oxidation and reduction processes. Each half-reaction is balanced for mass and charge, allowing for a clearer understanding of electron transfer. This method is particularly useful in complex reactions, as it simplifies the balancing process by focusing on individual components.

Basic Medium Balancing

When balancing reactions in a basic medium, hydroxide ions (OH-) are added to neutralize any hydrogen ions (H+) that may appear during the balancing process. This is essential because it ensures that the reaction remains in a basic environment, which affects the species present and their interactions. The addition of water may also be necessary to balance oxygen atoms.

Chromate and Hydrochromate Species

Chromate (CrO4^2-) and hydrochromate (Cr(OH)4^-) are important species in redox chemistry involving chromium. Chromate is a strong oxidizing agent, while hydrochromate is formed in basic conditions. Understanding the interconversion between these species is crucial for balancing reactions involving chromium, as it influences the overall stoichiometry and electron transfer.