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Ch.19 - Electrochemistry

Chapter 19, Problem 93

Given the following half-reactions and E° values, write a balanced equation for the formation of Mn2+ and MnO2 from Mn3+, and calculate the value of E° for this reaction. Is the reaction spontaneous under standard-state conditions?

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Hello in this problem, we are asked to provide the balanced equation and calculate the value of the standard cell potential for the formation of oxygen and water from hydro peroxide given the following half reactions. And the corresponding standard reduction potential values were asked to determine if the reaction is spontaneous under standard state conditions. Looking at the standard reduction potentials for the half reactions that we've been provided. The first one then that describes the reduction of hydro peroxide under acidic conditions to form water, we have a larger standard reduction potential than that for the second half reaction where we have oxygen being then reduced form hydro peroxide under acidic conditions. Therefore the larger and more positive standard reduction potential tells which which of these two will be the reduction half reaction. So this will be the reduction half reaction. Because we have a standard higher standard reduction potential. And the next reaction then once we write it in reverse, this will be our oxidation half reaction. We will rewrite these two Reactions, writing them in the correct direction. So we have our first reaction then is written as it's provided. So this is our reduction half reaction. And the standard reduction potential Is equal to 1.78V. And then the next reaction we write in reverse. So this will be our oxidation half reaction to have hydro peroxide. Then being oxidized to form oxygen gas And standard oxidation potential then is equal in magnitude but opposite and signed to the standard reduction potential. So this works out to negative 0.70V. We will combine our two half reactions, canceling those things that are the same on opposite sides. So we have two moles of hydrogen ions on the reacting side, and two on the product side. So those cancel. We have two electrons on the reactant and product sides. Those cancel. And then we combine what's left over. When we get to moles hydrant peroxide and goes to form oxygen, gas and water and our standard so potential. Then we combine our standard reduction potential and standard oxidation potential 1.78V -0.70V. So we get 1.08. Since this is positive and sign, It means then that the reaction will be spontaneous under standard state conditions. So, our reaction then we were told to write a balanced equation for the formation of oxygen and water from hydrant peroxide. So that's what we have here. The standard cell potential then is 1.08V. Given that it's positive in sign, it means we have then a spontaneous reaction under the standard state conditions. And this corresponds then to answer C. Thanks for watching. Hope this help