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. (a) What is E° for the reaction if the standard half-cell potential for the reduction of acetic acid to ethanol is 0.058 V?

Question

Accepted Answer

Hello. In this problem, we are told an experiment is conducted on an alternative source for the production of methanol. One way to determine if methanol is produced is through tight rations with acidic potassium bromate. The presence of methanol can then be determined since like Rome it is converted to chromium three and indicated by a color change from orange to green During the tight rations, methanol is oxidized formaldehyde as described in the following chemical reaction. If the standard half cell potential for the reduction formaldehyde, methanol is 0.110V, what is the standard cell potential for the reaction? Let's begin by considering the reaction that will take place at the cathode. We call reduction occurs at the cathode. We will have di chrome eight being reduced, perform chromium three. You can find these half reactions in a table of standard reduction potentials. Standard reduction potential for this reaction then Is equal to 1.36V. We then consider the reaction that's taking place at the anodes. Recall that oxidation occurs at the anodes. In this case we have methanol than being oxidized form formaldehyde. So in order to balance this reaction, we first begin with those elements other than hydrogen and oxygen. So we begin with the carbon. So the carbon is balanced as it's written, then we would add water to balance the oxygen, the oxygen is also balanced. So then we balance the hydrogen by adding hydrogen ions. We have four hydrogen is on the reacting side and two on the product side. So we'll add two hydrogen ions. To balance the hydrogen. We then need to balance the charge. It's neutral on the reactive side and we have a plus two charge on the product side. So we'll add two electrons to balance the charge. But then combine these two half reactions. To get the overall reaction. We see then that we have six electrons being gained. We need to have six electrons being lost. So that means then the reaction that's taking place at the an ode will multiply through by three. We see then that we can cancel our six electrons and we also have hydrogen in that both reacting to the product side. When we multiple apply through by three, we'll end up with six protons on the product side that will cancel with six on the reacting side and we're left with eight. Then we'll combine these two half reactions. So we get three moles of methanol reacting with micro mate. Under a city conditions perform three moles formaldehyde and two moles of chromium, three and seven waters. This is our overall reaction that's taking place. This matches what was given in the problem statement. Now, to find the standard cell potential, find the standard oxidation potential, we know that it is equal to but opposite signed to the standard reduction potential. We were told that the standard reduction potential for the reduction of formaldehyde or reduction of formaldehyde. Methanol is 0.110 volts. So the standard oxidation potential is negative 0.110 volts. So our standard cell potential, then we have the standard reduction potential plus the standard oxidation potential. This works out to then 1.36V -0.110V, which works out to 1.25V. So note that although we multiply through the reaction by three, this does not impact the standard oxidation potential. Since this is an intensive property, it's independent of the quality of our reacted products. And so the standard cell potential then for this reaction is 1.25V. Thanks for watching. Hope this helped.

Verified Solution

Key Concepts

Standard Electrode Potentials

Redox Reactions

Nernst Equation