You are on your dream vacation at the beach when a major storm knocks out the power for days. Your cell phone is dead, and you want to make a battery to charge it. You find the following materials in the beach house. blue stone algaecide for pools, which can be used to make a 1.0 M Cu 2+ solution alum in the kitchen, which can be used to make a 1.0 M Al 3+ solution aluminum foil, copper wire, and bologna, which can be used as a salt bridge. (d) An iPhone requires 5.0 V for charging. Can this battery charge the phone? Explain.

Welcome back, everyone. A photographer was on a trip to an exotic place to capture stunning landscapes. When the electrical system of his RV was knocked out. He tried to locate himself on the road using the Maps app on his tablet. But the tablet's battery was dead in his RV. He found safety matches that could use or that he could use to make a one molar solution of potassium chlorate. He bought his RV from an old photographer. So he was able to find a small amount of silver nitrate and made a one molar solution. He was wearing a silver ring and had a $100 American platinum eagle coin in his pocket. He used a paper towel dipped in salt solution as a salt bridge and the battery in the tablet requires five volts for charging. Can this battery charge this tablet? Because the safety matches have potassium chlorate que Cielo three recall that this is an ionic compound AK assault and will fully dissociate into the following ions will be formed potassium plus one as a product and our chlorate cielo +31 minus an ion. Take note that we're also told that the photographer uses silver nitrate and forms a solution of one molar concentration. And because silver nitrate is also an ionic compound ak assault, it's going to fully dissociate into our silver plus one cat ion and our nitrate and ion. So we need to pick which between which of these solutions would be our cathode or are an ode. Let's begin by recalling that for chlorate are first an ion. We can refer to our standard reduction potential table in our textbooks and see that it undergoes a reduction in which it forms as products. Chlorine, gas and water. This comes straight from our textbooks and in order to understand this reduction reaction, we need to balance out our atoms and charges. So notice that on the product side, we have only one mole of oxygen. Whereas on the reactant side, we have three moles of oxygen. In order to bounce that out, we're going to place a coefficient of two and we're rather a three, a coefficient of three in front of our water on the product side to give us three moles of water, which now also introduces six moles of protons on the product side. So we would go ahead and add also six moles of protons to the react inside to balance that out. Next. Notice that we only have one mole of chlorine on our reactant side where we have two moles of chlorine on the product side. So we're going to place a coefficient of one half in front of our chlorine gas on the product side, which is going to reduce our chlorine on the product side to just one mole to match with chlorine on the reactive side. Now, with our atoms balance, notice that we now have a net charge of positive six minus one. So a net charge of plus five on the reaction side and a net charge that is neutral on the product side. So to cancel out that plus five charge on the reactant side, we're going to expand our reactant side and add five electrons. So now we will have a net neutral charge on both sides of our reaction. Now, all of our atoms and charges are balanced and notice that in our textbooks for the following reduction and we know that it's a reduction since we have electrons on our react inside here. So ultimately, chlorate is gaining electrons for this reduction, we have a standard reduction potential value is not equal to 1.47 volts. And this comes from our textbooks recall that for a reduction reaction this will occur at the cathode. So we've already determined which electrode this solution will be present in. Now, we can assume that in our next reaction are silver catalon is going to likely make up are an ode. So let's show its reduction reaction first because we would begin with silver Catalan which will form solid silver. Ultimately, as a product, we need to cancel out that net charge of plus one on the reacting side. So we're going to add one electron on the reactant side. Now giving us a net neutral charge on both sides of our reaction. And recall that for the reduction potential of this reaction in our textbooks will see a value for n not equal to 0.80 volts. Now notice we see silver gaining one electron. So this is again for the reduction of silver catalon, but we already have our cathode solution for the reduction of chlorate. So we're going to need to flip this equation. And what we're going to flip to is solid silver on the react inside. Now, which is formed from silver caddy on gaining an electron Because we reversed our reaction. We're going to need to reverse the magnitude of our standard cell potential. So now we would have a standard cell potential equal to negative 0.80 volts instead of positive. And now we can understand that this reaction we've shown is an oxidation since it shows our silver releasing an electron as a product. And recall that our oxidation reactions occur at our an ode in our electric cell. So now we need to come up with an overall reaction. So taking our first half reaction, which will say is here, our second half reaction is our oxidation half. For our reduction half reaction, we have chlorate, an ion reacting with six moles of protons Plus five electrons. To produce 1/2 mole of chlorine gas Plus three moles of liquid water. And then for our second reaction for our an ode, we have solid silver which form silver caddy on And one electron releases one electron. So notice that in order to add these two reactions together to come up with an overall reaction, we're going to need to have matching electrons. So we're going to need to take the entire second equation and multiply it by a factor of five. This is now going to change this reaction to 5Molds of solid silver produces five moles of silver caddy on Plus five electrons. And now from here, we can add these together to take our or to get to our overall equation in which we want to cancel out the five electrons on the reactant side. In our first equation are reduction equation with the five electrons on the product side of our oxidation reaction. And everything else that we're left with, we will carry down that is going to leave us with our chlorate, an ion Plus five moles of solid silver plus six protons yields a half mole of chlorine gas Plus five moles of silver cat ions And then plus three moles of liquid water. Going back to the prompt notice that we're told that the photographer had a $100 American platinum eagle coin. This means that platinum or the coin rather will say the platinum coin serves as our electrode which is conducting the electricity. And we want to recall that platinum would make a great electrode because platinum is inert. So it's unreal active and this will prevent, will say that this therefore prevents platinum from reacting with, I'm sorry, that's wrong. So the platinum being a nerd is going to prevent our silver in the ring from reacting with our chlorine, which we see chlorine was chlorine gas was produced from our reduction of our chlorine an ion. And just so everything is visible. This is my note here. So this is why we see in our overall reaction, silver is instead reacting with chlorate but not chlorine, which again, chlorine is produced from our reduction of chlorate in our first have reaction. So now with this overall equation, we want to come up with a overall standard cell potential. Eh not. So recall that Bennett cell is equal to our self potential of our cathode plus our standard cell potential of our an ode. And so plugging in our values for our self potential of our cathode, we determined the value to be 1.47 volts plus our standard cell potential of our an ode for the oxidation of solid silver, which we determined to be negative 0.80 volts because we reverse the reaction to show the oxidation. So taking the sum here, we find that our cell potential S L is equal to 0.67 volts and going back to our prompt we're told that the battery on the tablet requires five volts. So because 50.67 volts is less than five volts that are required to charge the tablet, we can say that therefore, the photographer cannot charge his tablet. And so he cannot use the battery that he creates. And so for our final answer, we can confirm that the electric magnetic force E M F of the .67 volt battery is not enough to charge the tablet or to charge his tablet. So our final answer highlighted in yellow is this statement here and corresponds to choice D and the multiple choice to complete this example. I hope this made sense and let us know if you have any questions.

Ch.19 - Electrochemistry

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McMurry 8th Edition

Ch.19 - Electrochemistry

Problem 120d

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Chapter 19, Problem 120d

You are on your dream vacation at the beach when a major storm knocks out the power for days. Your cell phone is dead, and you want to make a battery to charge it. You find the following materials in the beach house. blue stone algaecide for pools, which can be used to make a 1.0 M Cu²⁺ solution alum in the kitchen, which can be used to make a 1.0 M Al³⁺ solution aluminum foil, copper wire, and bologna, which can be used as a salt bridge. (d) An iPhone requires 5.0 V for charging. Can this battery charge the phone? Explain.

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Identify the half-reactions for the copper and aluminum electrodes. The copper half-reaction is Cu^{2+} + 2e^- \rightarrow Cu, and the aluminum half-reaction is Al \rightarrow Al^{3+} + 3e^-.

Determine the standard reduction potentials for each half-reaction. The standard reduction potential for Cu^{2+} + 2e^- \rightarrow Cu is +0.34 V, and for Al^{3+} + 3e^- \rightarrow Al is -1.66 V.

Calculate the standard cell potential (E^\circ_{cell}) by using the formula E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}. Substitute the values: E^\circ_{cell} = 0.34 V - (-1.66 V).

Evaluate whether the calculated E^\circ_{cell} is greater than or equal to 5.0 V, which is the voltage required to charge the iPhone.

Conclude whether the battery can charge the phone based on the comparison of the calculated cell potential with the required voltage.

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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. In these cells, oxidation occurs at the anode, releasing electrons, while reduction occurs at the cathode, where electrons are accepted. Understanding how these cells function is crucial for determining if the materials available can create a viable battery to charge a device.

Standard Electrode Potentials

Standard electrode potentials are values that indicate the tendency of a species to be reduced, measured in volts. By calculating the difference in electrode potentials between the anode and cathode reactions, one can determine the overall voltage produced by the electrochemical cell. This is essential for assessing whether the battery can generate the required 5.0 V to charge the iPhone.

Salt Bridge Functionality

A salt bridge is a component of an electrochemical cell that maintains electrical neutrality by allowing the flow of ions between the two half-cells. It prevents the buildup of charge that would otherwise stop the reaction. In this scenario, using aluminum foil, copper wire, and bologna as a salt bridge is important to ensure the battery operates effectively and can sustain the necessary voltage for charging.