Determine whether or not each metal dissolves in 1 M HIO 3 . For those metals that do dissolve, write a balanced redox equation for the reaction that occurs. a. Au b. Cr

everyone in this example, we're asked whether copper metal will dissolve in one Moeller ionic acid. If yes, we need to write the balanced redox reaction that occurs upon dissolving our copper metal. So what we should recall and refer to is our reduction half reaction table, otherwise known as our standard electrode potential table. We're going to refer to this table and we're going to see that ionic acid oxidizes as the ion I date. And on our tables we would find the following reaction with I date, we're on our tables, we have plus six hydrogen ions plus five electrons. This will yield one half of our di atomic molecule iodine, I'm sorry, that should be a Q plus we have three moles of water. So this is all coming from our standard electrode potential table and this oxidation has a value according to our tables equal to 1.20 volts. When we refer to this table and finder adam copper, we would see that copper oxidizes as the two plus cat ion, where we would add two electrons here so that we form our copper metal. And according to our charts, this has a voltage equal to .34V. And so we would say that because this value is lower than our voltage for our oxidation of ionic acid, which had a value of 120V. We would say therefore copper will be dissolved in one moller iota acid. And that is because again it's lower on our standard electrode potential table and it only has a voltage of 10.34 volts. So our next step is to balance our electrons in both reactions so that we have the same amount for both of our reactions here. So we're gonna go ahead and take our first reaction. We're going to place it in brackets. And because we have five electrons here, we're going to multiply this entire reaction by a value of two. So just so it's visible, I'm just gonna move it over now. Just to be clear, we start out with our copper metal so we're just going to rewrite this reaction so that we have copper metal equal to or rather we can just say yields the copper two plus cat ion plus our two electrons. And so now it's clear that this is going to be the oxidation half reaction. And we recognize this because our electrons are on our product side here meaning we lost these two electrons to form our copper metal. So these electrons being on the product side means that this is our oxidation half reaction. And because we have two electrons here, we multiplied our first reaction here where we have our electrons on the reacting side, meaning that this was our reduction half reaction. We multiply this by two which would give us a total of 10 electrons. So we want 10 electrons for our oxidation half reaction, meaning we're going to place this in brackets and we're going to multiply this by five So that we can get 10 electrons for this half reaction as well. So in doing so we can rewrite these two half reactions. So for our first we would now have two moles of our I date an ion added to 12 moles of our H plus proton Plus 10 electrons now yields just one mole of ri date or sorry I died And this is added to now six moles of water. So this is our new reduction half reaction. And now for our copper we would now begin with five moles of our metal, copper which is yielded from five moles of our copper two plus cat ion plus our now 10 electrons. And now at this point we want to go ahead and cancel what is the same on opposite sides of our equations. And we would see that we have 10 electrons here on the react inside and then 10 electrons here on the product side. For our second reaction. So we can go ahead and actually cancel this out to yield are balanced redox reaction as a whole. And what this is going to give us is are bound three docks reaction where we should have now two moles of our I date an ion Plus our 12 moles of our proton h plus. Now coming from our second reaction, we have that 5Molds of our copper metal And now for our product side we're going to yield one mole of iodine Plus six moles of water and we're just gonna make more room here. And then for our last product we would have from our second reaction, the five moles of copper two plus. And so this would be our final answer here as our full balanced redox reaction. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below. So just to be clear what's highlighted in yellow is our balanced redox reaction. And we also can confirm our first answer is that copper will be dissolved in one moller I OTIK acid because it is lower on our standard electrode potential table than our I. O tick acid. So I hope that everything I explained was clear. If you have any questions, just leave them down below and I will see everyone in the next practice video.

Ch.19 - Electrochemistry

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Tro 4th Edition

Ch.19 - Electrochemistry

Problem 60

Chapter 19, Problem 60

Determine whether or not each metal dissolves in 1 M HIO₃. For those metals that do dissolve, write a balanced redox equation for the reaction that occurs. a. Au b. Cr

Verified step by step guidance

Identify the oxidation states and properties of the reactants. Gold (Au) is in its elemental form with an oxidation state of 0, and in HIO3, iodine typically has an oxidation state of +5.

Determine the potential for Au to be oxidized. Since Au is a noble metal, it is less reactive and has a higher reduction potential, meaning it is less likely to lose electrons and oxidize under normal conditions.

Consult the electrochemical series to compare the reduction potential of Au and the oxidizing power of HIO3. If the reduction potential of Au is higher than the oxidation potential of HIO3, Au will not dissolve.

If Au can be oxidized by HIO3, write the half-reactions for each species. For Au, the oxidation half-reaction would be Au → Au^3+ + 3e^-. For HIO3, the reduction half-reaction would involve the reduction of iodine.

Combine the half-reactions to form a balanced redox equation ensuring that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction. Adjust coefficients as necessary to balance atoms and charges.

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Key Concepts

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

Redox Reactions

Redox reactions involve the transfer of electrons between two species, resulting in changes in oxidation states. In these reactions, one species is oxidized (loses electrons) while another is reduced (gains electrons). Understanding redox processes is crucial for predicting how metals will react with acids, such as HIO3, and for writing balanced chemical equations.

Metal Reactivity

The reactivity of metals varies significantly, influencing their ability to dissolve in acids. Noble metals like gold (Au) are less reactive and often do not dissolve in dilute acids, while more reactive metals can readily react with acids to form metal ions. Knowing the position of metals in the reactivity series helps in predicting their behavior in acidic solutions.

Balanced Chemical Equations

A balanced chemical equation represents a chemical reaction with equal numbers of each type of atom on both sides of the equation. Balancing equations is essential for accurately depicting the stoichiometry of reactions, including redox reactions involving metals and acids. This ensures that the conservation of mass is maintained and allows for the calculation of reactants and products.