When suspected drunk drivers are tested with a Breathalyzer, the alcohol (ethanol) in the exhaled breath is oxidized to acetic acid with an acidic solution of potassium dichromate: The color of the solution changes because some of the orange Cr₂O₇^2- is converted to the green Cr³⁺ The Breathalyzer measures the color change and produces a meter reading calibrated in blood alcohol content. (b) What is the value of E for the reaction when the concentrations of ethanol, acetic acid, Cr₂O₇ are 1.0 M and the pH is 4.00?

Video transcript

Welcome back everyone. We're told that breathalyzers are used to test the amount of blood alcohol level in the exhaled breath of suspected drunk drivers. An acidic solution of potassium di chrome. It is usually used to oxidize alcohol, ethanol to acetic acid. In an experiment, an acidic solution of potassium permanganate was used rather than potassium di chrome. So we're given this overall equation here and we're told that potassium permanganate changes color from a deep purple to a colorless solution. After being reduced, the Breathalyzer can measure this change in color to get a reading of how much alcohol is in the blood, determine the potential of the reaction. If the concentrations of ethanol, acetic acid and permanganate as well as manganese carry on is 1. molar and the ph of the solution is 4.50. Were given the standard reduction potential of acetic acid equal to 0.58 volts. And the standard of reduction potential of per man getting equal to 1.51 volts. So let's begin by breaking down this overall reaction into two half reactions. So our first half reaction would show our five moles of ethanol, CH three ch 20. H, Which forms five moles of acetic acid. So we need to begin by balancing out our atoms, we have a total of Five times our one mole of oxygen. So we have five moles of oxygen on the reactant side and then we have five times two which would give us 10 moles of oxygen on the product side and so we need to bounce out oxygen on the reactant side by recalling that we can balance oxygen using water and so we would expand our react inside so that we would add another Five moles of oxygen by adding five moles of water to the reactant side. And so now you can see that we have five plus five, which would give us 10 moles of oxygen on the react inside. And then we would have the 10 moles of oxygen here, also on the reactant side. So now that our oxygen is balanced because we've introduced five moles of water to the reacting side, We've now changed our number of hydrogen for a count of hydrogen on the reactant side, we have five times three giving us 15 plus five times the subscript of two here giving us 15 plus 10. So that's 25 plus our five again times the single hydrogen here, which is 25 plus five. Giving us now 30 on the reacting side and then accounting for the 10 from the water. We have five times two. Giving us 10, we would have 30 plus 10, which would equal a total of 40 moles of hydrogen on the reactant side. So if we have 40 moles of hydrogen on the reacting side, we need to compare this to our product side. We're on our product side. We have five times three, which would give us 15 plus five times one, which would give us five. So that's 15 plus five which would give us 20. So we're missing 20 from the product side. And recall that we balance hydrogen using protons. So we would expand our product side and add a total of 20 moles of protons. So now that we have oxygen and hydrogen balanced as well as carbon, we can now focus on balancing out net charge on our reactive side. We have a net charge that is neutral. But on our product side we have a net charge of plus 20. So we need a neutral charge on our product side. And so just to make sure we have enough room. I'm just going to move this over To cancel that net charge of plus 20 on the product side, we're going to add 20 electrons so that we have a neutral charge on both sides of the reaction. Now recognize that because we have electrons on the product side, this tells us that this reaction occurs as an oxidation. And so for the oxidation of Ethanol, we would find in our standard reduction potential a value equivalent to 0.058V from our textbooks. And actually we can see that this is also given to us in our prompt here. So notice that our reduction potential is negative here. Well, we know that it's negative because our half reaction for ethanol is occurring as an oxidation. Whereas according to the prompts are reduction potential for ethanol Is for the reduction of ethanol. And so we just want to make this opposite meaning negative 0.058 as we have here because again, this is an oxidation instead of a reduction. And again, we know that this half reaction is an oxidation because electrons are on our product side. Now moving on to our second half reaction, We have four moles of permanganate, which forms formals of manganese two plus caddy on Beginning by bouncing out our moles of oxygen because our manganese atoms are balanced. We have a total of four times four, which would give us 16 moles of oxygen on the reacting side. We have no oxygen on the product side. So we're going to add 16 moles of water to our product side. Now with this edition of water, we've introduced hydrogen and we have a total of 16 molds or sorry, 16 times two. And so we would have 32 moles of hydrogen on the product side, which we need to bounce out on our reactant side. And so we would again balance hydrogen using protons. So we would add 32 moles and there should be a plus here, we would add 32 moles of protons to our react inside. Now with this edition of 32 moles of protons, we need to balance out net charge because all of our atoms are now balanced. We have a net charge of four times positive two. So that would be a net charge of positive eight on the product side. And we have a net charge on our React inside of positive And then -4. So that's 32 -4, which would give us a net charge of positive 28. So we're comparing a net charge of positive 28 to a net charge of positive 16. And so we would And sorry, I misspoke there were comparing a net charge of positive 28 for the reactive side to a net charge of positive eight on the product side. So we're going to decrease that higher charge on the reactant side By adding electrons to the react inside. And so we would specifically want to add a total of 20 electrons giving us a net charge of plus eight on both sides of the reaction. And so all of our atoms are balanced. And because we see that electrons are on the react inside, in this case we can label this second half reaction as a reduction. And because in the prompt were given the standard reduction potential for permanganate as 1.51V. Which actually ends up being exactly how our half reaction occurs as a reduction. We can say that our reduction potential under standard conditions is definitely equal to positive 1.51V. So we won't change the sign here. Now we're going to move on now that we have our half reactions balanced and their noted reduction potentials. We want to calculate our total cell potential under standard conditions. E degrees cell which we should recall is calculated by taking our cell potential of our an ode, plus our self potential of our cathode. So recall that our reduction occurs at the cathode and our oxidation reaction occurs at the anodes. So we would have the sum of negative 0.058V added to 1.51V. And this would give us a total standard cell potential of 1.452V. So now with this total standard cell potential, we're going to now recall for the next step of this prompt the Nerdist equation to determine the self potential of our reaction under nonstandard conditions, which is what is described at the end of our prompt recall that the nearest equation is where our cell potential under nonstandard conditions, which is represented as S. L. Is equal to our cell potential under standard conditions, not sell minus 0.592 volts divided by n electrons transferred, multiplied by the log of Q. Our reaction quotient and recall that Q. Is equal to the concentration of our Aquarius products over our concentration of our Aquarius react ints. So we're going to be referring to our overall equation given in the prompt where we have again, five moles of ethanol, ch three ch 20. H Plus four moles of permanganate Plus 12 moles of protons yields five moles of acetic acid Plus four moles of manganese Catalon and sorry for most of manganese catalon Plus 11 moles of water. And this is a plus here. And so because we know our reaction quotient is our concentration of products over reactant. We would have our concentration of our various products being our acetic acid. Or sorry? Yes, our acetic acid, ch three C. 02 H. Which has a coefficient of five, which will raise as a power of five here multiplied by a concentration of manganese carry on, which has a coefficient of four, which we raise as a power of four divided by our concentration of our acquis reactant being our ethanol. So ch three ch 20. H. Which has a coefficient of five, which we raise as a power of five. Multiplied by our second reactant concentration of permanganate which has a coefficient of four, which we raise as a power of four. And then we also have our protons which are Aquarius and so on a reaction side. This is also in our denominator. So we multiplied by the concentration of age plus which has a coalition of 12 which we raised to a power of 12. Now recall that from our prompt we are told that our concentration of ethanol or sorry, our concentration of acetic acid as well as our concentration of manganese catalon are equal to each other. This is also equal to our concentration of ethanol and these are also equal to the concentration of permanganate And this entire or all of these concentrations as a whole from the prompts are equal to 1.5 older. This is given to us so we know all of the concentrations for those three agents. Which leaves us with figuring out what our concentration of our protons are. And so recall that we can find this by taking 10 to the negative value of our ph which were actually given in our prompt. So we can say we have 10 to the negative ph given in the prompt as 4.50. And this gives us our concentration of protons equal to a value of three point sorry three 0.16228 times 10 to the negative five molar, allowing us to now solve for our quantity of our reaction quotient Q. Because we know all our concentrations. So we would plug in 1.5 molar raised to the fifth power. Multiplied by 1.5 Mueller, raised to the 4th power, Divided by 1.5 Molar raised to the fifth power, multiplied by 1.5 Molar raised to the fourth power and then multiplied by three points 1622, 8 times 10 to the negative 5th power moller. And so with all of these concentrations plugged in, we're going to find that our value for Q. is equal to one times 10 to the positive 53rd power. And just to be clear, this is raised to a coefficient or power rather of 12 because we have a coefficient of 12 in front of our protons. So now that we have our value for the reaction quotient, we can calculate our cell potential under nonstandard conditions using our new ERnst equation. And so we would say that our cell potential under nonstandard conditions, S L. Is equal to our value for the self potential under standard conditions which we determined initially in our calculation of 1.452 volts. This is subtracted from 0.592 volts divided by our electrons transferred. So in our overall or rather in our half reactions, we had a transfer of 20 electrons in both of our half reactions. And so are electrons transferred. N Is equal to 20. So our denominator is 20 electrons. And then this is again multiplied by the log of our reaction quotient, which is one point or sorry, which is one Times 10 to the 53rd power. So we don't actually need to write electrons here, but N is equal to 20 electrons transferred. So now let's go ahead and simplify this in our calculators. And what we would have is our cell potential under nonstandard conditions is equal to 1.452V subtracted from when we divide the parentheses here, we get a result of 0.15688. So that is from the division here in our parentheses multiplied by the log of our reaction quotient Q. And so we would be left with units of volts and taking the difference between our volt units, we find ourself potential under nonstandard conditions equal to a value of 1.295 12 volts. Now recall from the prompt, we have a minimum amount of sig figs being three sig figs surrounding this off 2366. We would just have one point 29V. And so this would be our final answer as the self potential under nonstandard conditions for our experiment, this corresponds to choice a in the multiple choice, I hope everything argued was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.