A mercury battery uses the following electrode half-reactions: (c) What is the effect on the cell voltage of a tenfold change in the concentration of KOH in the electrolyte? Explain..

Video transcript

Welcome back everyone. We need to consider the following electrode half reaction of an alkaline battery. How does a five fold increase in the concentration of sodium hydroxide in the electrolyte affect the cell voltage and explain So, note that we have the following reactions where we have two moles of manganese dioxide solid reacting with one mole of water and releasing two electrons, where we form one mole of manganese, three oxide solid and two moles of hydroxide were given a reduction potential equal to 20.15 volts for this reaction. And then for the second reaction, we have zinc oxide solid reacting with water, releasing two electrons to form solid zinc and two moles of hydroxide with a corresponding reduction potential of negative 1.25 volts. Now we need to identify which of these is our anodes or cathodes reaction. So recall that the lower our standard cell potential that will correspond to an oxidation reaction, which we should recall will always occur at our an ode of our voltaic cell. And that means that on the converse, the higher the standard reduction potential will correspond to a reduction reaction, which we recall occurs at the cathode of our voltaic cell. And so that means because we see we have a cell potential of or reduction potential of 0.15 volts, which is greater than negative 1.25 volts. We can consider this first reaction as our reduction, meaning that this reaction occurs at our cathode. And then our second reaction we can label as an oxidation because it has a more negative self potential occurring at our an ode. We can also observe that both of these reactions have matching number of electrons. So we can easily add them up. But we want to make sure that our oxidation reaction has our electrons on the product side. So let's take this second half reaction and flip it so that we have one mole of solid sink plus two moles of hydroxide Yields one mole of zinc oxide solid Plus one Mole of Water. And then those two electrons that are released and so adding this to our first half reaction we have for the first half reaction, two moles of manganese dioxide solid Plus our one mole of liquid water Plus our two electrons yields one mole of manganese three oxide Solid Plus two moles of hydroxide. And so adding these together, we can begin by canceling out our electrons in our product side of our oxidation here and then canceling out the electrons on the react inside of our reduction reaction. We can also cancel out from the product side of our oxidation reaction, The water, the one mole of water and then the one mole of water on the reactive side of our reduction reaction. And we can also get rid of the two moles of hydroxide in both reactions. So everything that we're left with will give us our overall reaction being one mole of solid zinc Plus two moles of manganese dioxide solid yields one mole of zinc oxide solid and one mole of manganese, three oxide solid. Now, with this overall reaction, we can observe that we don't have any sodium hydroxide present. And so a five fold increase of sodium hydroxide will likely not affect our cell voltage. Also recognize that if we want to calculate the overall self standard cell potential here, we would recall the following formula where our standard cell potential is equal to 0.592 volts divided by our number, number of electrons transferred n multiplied by the log of our equilibrium constant for the reaction. Kay now we don't know our equilibrium constant so we would just need to figure that out. But overall to complete this example and answer our prompt, we can confirm that A 5-fold increase in our concentration of sodium hydroxide will not affect the cell voltage. And we can confirm that this is because it's not present in the overall reaction. And so this entire statement is our final answer corresponding to choice. Be in the multiple choice. I hope that everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video