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 FeO₄^2- 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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Hello. In this problem, we are told the ordinary alkaline battery. Half reactions can be written as shown below. Were asked to consider a hypothetical battery composed of the same an ode as the ordinary alkaline battery and a cathode where the cobalt oxide iron from sodium cobalt oxide is reduced to solid cobalt three hydroxide in a basic solution. Because the hypotheticals battery's cathode can provide more charge per unit mass than an ordinary alkaline battery of the same size and weight. It should last longer compare quantitatively the amount of columns of charge released when 15 g of sodium cobalt oxide is reduced to cobalt three hydroxide and the amount of columns of charge released when g of manganese dioxide is reduced to manga night. So we've been provided the half reaction for the reduction of manganese dioxide. Now this right the half reaction for the reduction of our cobalt oxide ion to cobalt three hydroxide, recall, the first thing we want to ensure is that everything is balanced besides that of hydrogen and oxygen. So we would first consider the cobalt. So in this case the cobalt is balanced. So now we can go about balancing the oxygen. We do that by adding water. We have four oxygen's on the reactant side, three on the product side. So we need another water to balance the oxygen's. Now that we've balanced the oxygen's we need to balance the hydrogen. We have five hydrogen on the product side, none on the reacting side and so well balanced that by adding five hydra nines. We then need to balance the charge. So it's neutral on the product side. We have a net three plus charge on the reactant side. So we're gonna add three electrons to the reactant side. We are told in this problem that this occurs in a basic solution. So we're gonna neutralize our hydrogen lines by adding an equal amount of hydroxide ions and we need to do that to both sides. So the hydra nines and the hydroxide ions then will combine to form five waters. And we can then eliminate the water on the product side. And then we remove one of the waters from the reacting side. Rewriting the reaction. Then we have three electrons plus four waters plus our cobalt oxide ion Close to Form Cobalt three Hydroxide. And we have five hydroxide lines. Now we are going to calculate the Amount of columns have charged released from 15 g of the sodium cobalt oxide is reduced. So we have our 15 g bar sodium oxide will convert that to molds using its smaller mass. It has more mass. 168.910 g. And then in every one mole Of this sodium oxide, you have one mole of our cobalt oxide ion. And then looking at our reduction half reaction that we wrote, we see that we have three moles of electrons for every one mole of our cobalt oxide iron. And then we're going to make use of a Faraday to convert from moles of electrons to columns. One more electron. So the charge 96485. Making sure units cancel. So we have grams of sodium cobalt oxide cancels moles of sodium cobalt oxide cancels our molds of our cobalt oxide. Iron cancels our moles of electrons cancel. And we're left with columns and this works out to 25,700 bombs. And then we're going to now calculate the Charge released from 15 g of manganese dioxide is being reduced. So we have 15 g uh manganese dioxide. We will convert that to molds using a smaller mass As Molar mass of 86.937 g. Using the reduction reaction, we find that for every one mole of manganese dioxide we have one more electrons make use again Our charge on one mole of electrons you can get from molds to columns and so our grams and manganese dioxide cancels, molds of manganese dioxide cancels moles of electrons cancels. And this works out to 16,600. So we have found the charge released when cobalt oxide iron is being reduced and then the amount of charge for when manganese dioxide is being reduced. And we see that the amount of columns that is released is greater for the reduction of the sodium cobalt oxide. This all then corresponds to answer C Thanks for watching help this help

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Key Concepts

Half-Reactions in Electrochemistry

Molar Mass and Stoichiometry

Coulombs and Charge Calculation