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Ch.19 - Electrochemistry

Chapter 19, Problem 163

Gold metal is extracted from its ore by treating the crushed rock with an aerated cyanide solution. The unbalanced equation for the reaction is (b) Use any of the following data at 25 °C to calculate ∆G° for this reaction at 25 °C: Kf for Au(CN)2- = 6.2 x 10^38, Ka for HCN = 4.9 x 10^-10, and standard reduction potentials are

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Hello. In this problem, we are told that silver is extracted from its ore vehicle leaching process with cyanide iron in that ore shown below is the overall reaction. We are asked to use the given data to calculate the change in Gibbs free energy for the above reaction at 25°C. In the first two reactions. Then we have the equilibrium constant. Associate with those reactions. So we can relate the change and gives free energy to. Our equilibrium constant is equal to negative of the gas constant times, temperature times the natural log of the equilibrium constant. The last two reactions then involved electrochemical cell potential. So we can relate the electrochemical cell potential to the gibbs. Free energy is equal to negative end, which is the most of electrons that are transferred. F is the fair day. Andy then is our electrochemical cell potential. We also want to make use of the N. R. Gas constant Which is 8. Kill jules per kelvin, 8.3 and four times 10 negative three killed joules per kelvin. And our temperature is 25 degrees Celsius. So we'll convert that to Kelvin. So it works out to 298.15 Kelvin. So beginning with our first reaction then that involves silver ions. The change in gibbs free energy then is going to be equal to the negative of our gas constant times are temperature Times in the natural log of our equilibrium constant, which is three times 10 to the 20. So our units of Kelvin cancels. So we're left with kill jules. This works out to negative 116. 771 Gold Jewels. For the next one that involves water, we have the change in gibbs. Free energy is equal to again, the negative of our gas constant, times are temperature Times Natural Log of Our Equilibrium Constant, which is one times 10 to the -14. And so our units of kelvin, oops, sorry, this should be for kelvin, sorry, units of kelvin cancels. And we're left with then till Jules. This works out to then 79.9077. Kill jules. The next one finding the change that gives for energy for the reaction involving oxygen, we have four moles of electrons That are being gained. And so we have negative four moles have electrons times the Faraday, which is 96,485 g per mole of electrons Then have our voltage which is 1.229V. We call that one jewel Is one volt. And we will convert our jewels to kill jules. So checking that our units cancel. So we have most of electron cancels, cancels volts cancels and jewels cancels were left with kill jules. This works out to negative 474.3 to 0 fill jewels. And then the last one involving silver ions. We have Then in this reaction we have one mole of electrons that's being gained to have negative one mole electrons. Time is fair day. The charge on a mole of electrons. We have a voltage which is 0.800V. We have one jewel per boat. We'll convert our jewels to kill jules. So making sure units cancel multi electrons cancel, coolants, cancels, volts cancels, and jewels cancels. And we're left with kill jules. This then works out to negative 0.1 88 killed jules. That's now right. Are associated change in Gibbs, free energy with each of these reactions. So for the first one then we had negative 116.8771. Kill jules. The next one, it was 79. 3rd 1, we found it was negative 474.320. And for the last one then worked out to negative 77.188 gold jewels. And now we're going to combine these reactions to get the overall reaction. That was given in the problem statement shown here, is that overall reaction that was provided. So this will be our guide as to how we make use of the equations above, we will determine if they're written in the direction that they're needed. If not, we will reverse them. And then based on the number of um of each of those different um reactant or products, we will multiply through by the proper coefficient. So when we look at the first one, we see then that are complex sign between silver and cyanide is on the product form. And the reaction that were provided and it's also on the product side in the reaction that we're after. So this is written in the direction we wanted. However, in our reaction that we're after, then we have formals of our complex between silver and Sinai. So we're gonna multiply the whole reaction through by four. When we do that, we then also have to multiply the change in gibbs free energy by four. So we will have four moles of silver Ryan's Plus eight moles of cyanide thing goes to form for moles the silver cyanide. And we're going to multiply the gibbs free energy value by four. So we'll have four times to get 116.877 killed jewels for the next reaction then. So we have water and we have hydroxide ions. And so it is written in the direction that we need it. So we have water on the reaction side in both cases, and hydroxide ions on the product side, we see that we have four moles of hydroxide ions. So we will also multiply this reaction by four. So we get four moles of water formals of hydrogen ions and four moles of hydroxide ions. And again, we need to multiply. Then our change in gibbs free energy by four as well. So four times 79.9077 killed jules for the next reaction, then we have oxygen and we see that oxygen is on the reactant side for both reactions. So again this is written in the direction that we needed. And this is also then um the crap number of oxygen molecules. So we just will rewrite this equation. We have oxygen gas plus four protons, plus four electrons. First form two waters And so we have the same tip free energy value as right above 474.3-0 gold jewels. And then the last reaction we have silver ions and silver medal. So the silver medal is on the product side in the reaction that's provided and it's on the react inside on the reaction that we are after. And we also have four moles of silver. So we will reverse the reaction. Which means we change the sign Of the Gibbs. Free energy from negative 77.188 children's to positive and will also multiply it through by four. So right the reverse reaction and multiply that four. They get four silver medal, those two form for silver ions plus four electrons. And again the change it gives for energy. Then we're gonna multiply three x 4. And we have changed the sign to the positive 77. kill jules. And now we want to combine all these reactions and are changing gifts for energy And verify that we get the reaction that is shown above it. So this reaction here. So when we look, we see that we have four silver ions on the reactant side and four on the product side, we have four electrons on the reactant side and we have four on the product side. We have for hydra nines on the reactant side, and four on the product side. And then we have four water molecules on the reactant side, and we have two on the product side. So we'll cancel two of those and then we'll combine what's left over to verify that we come up with the reaction that we were after. So we have four moles of silver. We have eight moles The cyanide, we have one mole oxygen gas. We have then in two waters. And this all then goes to form for of our silver complex sign with cyanide. And we also have four hydroxide. So these do some to the equation that we're trying to find the change in gibbs free energy for. And so we then we'll just add up to do the math for each of the change in gibbs free energies for each one of these steps and combine them to come up with the overall gifts for energy for the reaction. When we do that, this then works out to the change in Gibbs free energy is equal to negative 313 killed jules. And so this value for the change and gives you energy for that overall reaction corresponds Then to answer D. Thanks for watching Hope. This helps.
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Textbook Question

The half-reactions that occur in ordinary alkaline batteries can be written as In 1999, researchers in Israel reported a new type of alkaline battery, called a 'super-iron' battery. This battery uses the same anode reaction as an ordinary alkaline battery but involves the reduction of FeO42- ion (from K2FeO4) to solid Fe(OH)3 at the cathode. (a) Use the following standard reduction potential and any data from Appendixes C and D to calculate the standard cell potential expected for an ordinary alkaline battery:

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Textbook Question

The half-reactions that occur in ordinary alkaline batteries can be written as In 1999, researchers in Israel reported a new type of alkaline battery, called a 'super-iron' battery. This battery uses the same anode reaction as an ordinary alkaline battery but involves the reduction of FeO42- ion (from K2FeO4) to solid Fe(OH)3 at the cathode. (b) Write a balanced equation for the cathode half-reaction in a super-iron battery. The half-reaction occurs in a basic environment.

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Textbook Question

The half-reactions that occur in ordinary alkaline batteries can be written as In 1999, researchers in Israel reported a new type of alkaline battery, called a 'super-iron' battery. This battery uses the same anode reaction as an ordinary alkaline battery but involves the reduction of FeO42- ion (from K2FeO4) to solid Fe(OH)3 at the cathode. (c) A super-iron battery should last longer than an ordinary alkaline battery of the same size and weight because its cathode can provide more charge per unit mass. Quan-titatively compare the number of coulombs of charge released by the reduction of 10.0 g K2FeO4 to Fe(OH)3 with the number of coulombs of charge released by the reduction 10.0 g of MnO2 to MnO(OH).

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Textbook Question
Consider the redox titration of 100.0 mL of a solution of 0.010 M Fe2+ in 1.50 M H2SO4 with a 0.010 M solution of KMnO4, yielding Fe3+ and Mn2+. The titration is carried out in an electrochemical cell equipped with a platinum electrode and a calomel reference electrode consisting of an Hg2Cl2/Hg electrode in contract with a saturated KCl solution having [Cl-] = 2.9M. Using any data in Appendixes C and D, calculate the cell potential after addition of (a) 5.0 mL, (b) 10.0mL, (c) 19.0 mL, and (d) 21.0 mL of the KMnO4 solution.
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Textbook Question

We've said that the +1 oxidation state is uncommon for indium but is the most stable state for thallium. Verify this statement by calculating E ° and ΔG ° (in kilojoules) for the disproportionation reaction

3 M+1aq2S M3+1aq2 + 2 M1s2 M = In or Tl

Is disproportionation a spontaneous reaction for In+ and/orTl+? Standard reduction potentials for the relevant halfreactions are

In3+1aq2 + 2 e- S In+1aq2 E° = -0.44 V

In+1aq2 + e- S In1s2 E° = -0.14 V

Tl3+1aq2 + 2 e- S Tl+1aq2 E° = +1.25 V

Tl+1aq2 + e- S Tl1s2 E° = -0.34 V

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