Heart pacemakers are often powered by lithium–silver chromate 'button' batteries. The overall cell reaction is 2 Li(s) + Ag 2 CrO 4 (s) → Li 2 CrO 4 (s) + 2 Ag(s) (a) Lithium metal is the reactant at one of the electrodes of the battery. Is it the anode or the cathode?

Welcome back everyone in this example, we're told that everyday household items such as flashlights and radios use alkaline batteries. The overall reaction in the battery is as follows where solid zinc reacts with two moles of manganese, four oxide in an equilibrium reaction to form zinc oxide, solid and manganese manganese three oxide. We're told that at one of the battery's electrodes, zinc metal is a reactant. Which of the below statements is a true statement. So we need to see housings oxidation number has changed in our reaction. So recall that any substance in its standard state is going to have an oxidation number of zero. So zinc solid is in its standard state and will have an oxidation number of zero. Now on our product side, it's in a poly atomic compound. So we need to solve for zinc oxidation number as X. Here we're going to recall that oxygen always has an oxidation number of negative two. And overall we have a neutral compound, meaning we can sulfur for sinks oxidation number by saying X plus negative two is equal to zero, meaning that the oxidation number of zinc should be positive too. And so this is going to be zinc oxidation number on the product side. And we can see that we went from an oxidation number that was neutral to a more positive oxidation number as plus two, meaning we have an increase in oxidation number. And so we would say that therefore the reaction occurs as an oxidation And recall that any reaction that occurs as an oxidation will occur at the anodes of our electro voltaic cell. And so to complete this example, we would say that the only true statement is going to be statement B, which states that zinc solid is oxidized at the anodes. And this makes sense because we should recognize that. To go from that neutral oxidation number two, A charged oxidation number of plus two for our zinc on the product side are zinc solid had to lose two electrons. And recall that oxidation means we lose electrons. And so again, because this reaction occurs as an oxidation, it occurs at the and not have cell. And so this is going to be our only true statement as the final answer. Now we can double check this answer by referencing our alkaline batteries from the prompt where our first reaction should be sync oxide to form solid zinc. I'm sorry zinc oxide is also a solid where our first step is to balance out our atoms and because we have oxygen on the reactant side, we want to recall that to balance out oxygen, we use hydroxide. So we're going to add hydroxide to the product side where we now have introduced hydrogen by adding hydroxide. And so we're going to recall that to balance out hydrogen, we use water. So we're going to add water to the reactant side and this adds two moles of our oxygen and two moles of our hydrogen. So to bounce out the oxygen and hydrogen, we're going to place a coefficient of two in front of our hydroxide And hydroxide recall has a -1 and ion charge. So we also stated that we have a loss of two electrons. And so we're going to release or rather in this case we're going to just add it on the reactive side here since we're starting off with the solid zinc oxide. So these two electrons are released in our oxidation step and we want to recall that this oxidation reaction has a electrode cell potential equal to a value of negative 1.28 volts, which is negative and should occur at our note since this reaction occurs as an oxidation. So next we want to look at according to the prompt our second alkaline battery, which is our manganese for oxide. So we have M N. 02. We're going to produce manganese three oxide. I'm sorry, this is a solid label. So we're going to begin by bouncing out our atoms manganese by placing a coefficient of two in front of manganese on the reactant side, which is going to change our number of oxygen to now for most of oxygen. So we're going to bounce out oxygen by recalling again, we use hydroxide to bounce out oxygen adding in this hydroxide, we're going to need to place the coefficient of two here so that we have a total of six oxygen's But now that we have hydroxide which introduced hydrogen to our equation, we're going to need to recall to balance out hydrogen, we use water. So we're going to add one mole of water on our reactant side and making more room because we need to show our transfer of electrons. We're going to recall that in this reduction step. We're going to be releasing a total of two electrons, which we know because we'll recognize that our oxidation number of manganese went from an oxidation number of negative two to an oxidation number of negative three, all in reference to just our manganese adam. And so this is going to occur as a reduction because we have a decrease in our oxidation number to a more negative value here for manganese. And so this reduction of this manganese two or four oxide has a electro potential value of positive 0.15 volts which will occur at our cathode of our electro voltaic cell. And because we see that our oxidation reaction is consistent with occurring at the anodes for our zinc solid compounds, we can see that our correct answer as B definitely makes sense. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below. And I'll see everyone in the next practice video

Ch.20 - Electrochemistry

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Brown 15th Edition

Ch.20 - Electrochemistry

Problem 79a

Chapter 20, Problem 79a

Heart pacemakers are often powered by lithium–silver chromate 'button' batteries. The overall cell reaction is 2 Li(s) + Ag₂CrO₄(s) → Li₂CrO₄(s) + 2 Ag(s) (a) Lithium metal is the reactant at one of the electrodes of the battery. Is it the anode or the cathode?

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Identify the role of lithium in the cell reaction: \(2 \text{Li}_{(s)} + \text{Ag}_2\text{CrO}_4_{(s)} \rightarrow \text{Li}_2\text{CrO}_4_{(s)} + 2 \text{Ag}_{(s)}\).

Determine which species is oxidized and which is reduced. Lithium goes from \(0\) to \(+1\) oxidation state, indicating oxidation.

Recall that oxidation occurs at the anode. Therefore, the electrode where lithium is oxidized is the anode.

Understand that the anode is where the oxidation reaction takes place, and electrons are released.

Conclude that since lithium is oxidized, it is the anode in this electrochemical cell.

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Key Concepts

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

Electrochemical Cells

Electrochemical cells convert chemical energy into electrical energy through redox reactions. They consist of two electrodes: the anode, where oxidation occurs, and the cathode, where reduction takes place. Understanding the flow of electrons and the reactions at each electrode is crucial for analyzing battery function.

Oxidation and Reduction

Oxidation is the loss of electrons, while reduction is the gain of electrons. In the context of batteries, the anode undergoes oxidation, releasing electrons that flow through the circuit to the cathode, where reduction occurs. Identifying which species is oxidized or reduced helps determine the roles of the electrodes.

Lithium in Batteries

Lithium is commonly used in batteries due to its high electrochemical potential and light weight. In the given reaction, lithium metal is oxidized to lithium ions at the anode, providing a source of electrons. This characteristic makes lithium an essential component in the design and function of lithium-based batteries.