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Ch.19 - Electrochemistry

Chapter 19, Problem 158a

Consider the redox titration (Section 4.13) of 120.0 mL of 0.100 M FeSO4 with 0.120 M K2Cr2O7 at 25 °C, assuming that the pH of the solution is maintained at 2.00 with a suitable buffer. The solution is in contact with a platinum electrode and constitutes one half-cell of an electrochemical cell. The other half-cell is a standard hydrogen electrode. The two half-cells are connected with a wire and a salt bridge, and the progress of the titration is monitored by measuring the cell potential with a voltmeter. (a) Write a balanced net ionic equation for the titration reaction, assuming that the products are Fe3+ and Cr3+.

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Hey everyone, we're told that a redox situation of 100 and 50 mL of 0.110 moller chroma sulfate with 0.110 moller potassium permanganate was performed at 25 degrees Celsius with its ph being maintained at 2.50. Using a suitable buffer. One half of an electrochemical cell is made up of the solution in contact with a platinum electrode. The other half cell is a standard hydrogen electrode. The two half cells are joined by a wire and a salt bridge. And the cell potential is measured with a volt meter to track the tight rations progress if the expected products are chromium with a plus three charge and manganese with a plus two charge. What is the net ionic equation for the thai tradition reaction? So they've given us a lot of information here. First we have a ph of 2.50, which means that we are in an acidic environment and we have our chroma sulfate and this is going to become our chromium with a plus two charge plus our sulfate ion. Next, we were told that we had potassium permanganate and this dissociates into our potassium ions plus our permanganate ions. So our half reactions Are going to be our chromium with a plus two charge and this is going to become a chromium with a plus three charge. And we also have our permanganate and this will become our manganese ion with a plus two charge. Looking at our half reactions, it looks like we need to balance our oxygen and our hydrogen first and we can do so by adding water. And in order to balance out our oxygen's will have to add a coefficient of four prior to our water. But now we need to balance out our hydrogen and we can do so by adding our hydrogen ions into our react inside. And in order to balance this out we would have to add a coefficient of eight prior to our hydrogen ion. Now our hydrogen and oxygen are completely balanced. So let's go ahead and balance out our charges. Now for our chromium we have to add one electron in our product side to balance this out. Now, for our Other half reaction we have to add five electrons in our react inside to balance this out. Now to balance out our half reactions with one another, we have to multiply our first half reaction by five so we can get those electrons to cancel out with one another. So when we combine our half reactions we get our permanganate ion Plus eight of our hydrogen ions Plus five of our chromium with a plus two charge. And in our product side we get five of our chromium with a plus three charge Plus our manganese with a plus two charge Plus four of our water. And this is going to be our final answer. Now I hope that made sense. And let us know if you have any questions
Related Practice
Textbook Question

The reaction of MnO4- with oxalic acid (H2C2O4) in acidic solution, yielding Mn2+ and CO2 gas, is widely used to determine the concentration of permanganate solutions. (a) Write a balanced net ionic equation for the reaction.

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Textbook Question

The reaction of MnO4- with oxalic acid (H2C2O4) in acidic solution, yielding Mn2+ and CO2 gas, is widely used to determine the concentration of permanganate solutions. (d) A 1.200 g sample of sodium oxalate (Na2C2O4) is dissolved in dilute H2SO4 and then titrated with a KMnO4 solution. If 32.50 mL of the KMnO4 solution is required to reach the equivalence point, what is the molarity of the KMnO4 solution? .

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Textbook Question
A concentration cell has the same half-reactions at the anode and cathode, but a voltage results from different concentrations in the two electrode compartments. (b) A similar cell has 0.10 M Cu2+ in both compartments. When a stoichiometric amount of ethylenediamine (NH2CH2CH2NH2) is added to one compartment, the measured cell potential is 0.179 V. Calculate the formation constant Kf for the complex ion Cu(NH2CH2CH2CH2)22+. Assume there is no volume change.
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Textbook Question
Consider a galvanic cell that utilizes the following half-reactions:

(b) What are the values of E° and the equilibrium constant K for the cell reaction at 25 °C?
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Textbook Question
The nickel–iron battery has an iron anode, an NiO(OH) cathode, and a KOH electrolyte. This battery uses the follow-ing half-reactions and has an E° value of 1.37 V at 25 °C. (b) Calculate ∆G° (in kilojoules) and the equilibrium con-stant K for the cell reaction at 25 °C.
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Textbook Question

Experimental solid-oxide fuel cells that use butane (C4H10) as the fuel have been reported recently. These cells contain composite metal/metal oxide electrodes and a solid metal oxide electrolyte. The cell half-reactions are (b) Use the thermodynamic data in Appendix B to calculate the values of E° and the equilibrium constant K for the cell reaction at 25 °C. Will E° and K increase, decrease, or remain the same on raising the temperature?

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