A voltaic cell utilizes the following reaction: 4 Fe2+1aq2 + O21g2 + 4 H+1aq2 ¡ 4 Fe3+1aq2 + 2 H2O1l2 (b) What is the emf of this cell when 3Fe2+4 = 1.3 M, 3Fe3+4= 0.010 M, PO2 = 0.50 atm, and the pH of the solution in the cathode half-cell is 3.50?

Question

Accepted Answer

Welcome back everyone. We need to calculate the E M. F. For the voltaic cell with the reaction where we have four moles of chromium Catalan reacting with oxygen gas and four moles of hydrogen hydride caddy on to produce four moles of our chromium three plus catalon and two moles of liquid water. We're told are pressure for oxygen. It's 20.75 A t. M's are concentration of chromium Catalan is 1.7 moller and our concentration of chromium three plus Catalan is 30.12 molar. Were also given the Ph at the Cathode have cell equal to 3.25. Now, what we want to recognize based on each of these or the pressure. The concentrations and ph given is that we are at nonstandard conditions. So we want to recall that standard conditions Are going to be when we have a concentration of one molar, a ph equal to seven temperature being 25 degrees Celsius and a constant or pressure rather equal to 1 80 M. That would justify standard conditions for R E M F. Which we would recognize is also referred to as our self potential. Being e degree sell for standard cell potential. Now because we are at nonstandard conditions, we need to calculate what this E M F. Or sl So self potential at non standard conditions would be meaning we need to recall our nursed equation where we would say that our self potential at nonstandard conditions is equal to our self potential at standard conditions subtracted from the quotient where we take 0. 92 volts divided by our electrons transferred N, which is then multiplied by the log of our reaction quotient Q. And recall that our reaction quotient includes our concentration of our products over our reactant, which will only be our various substances. So what we first need to accomplishes writing out our half reactions based on our given reaction here. So for our half reaction, we'll begin with chromium, we have our chromium to Pluskat ion. So that's a queue here which produces as a product, our chromium three plus cat ion. And we gained that three plus charge because we release an electron here on the product side. We then have our second half reaction where we begin with oxygen gas and our four moles of our hydride caddy on. We're on the product side. We produce two moles of our water. Now, looking at the second half reaction, we've written because we have formals of hydride that gives us an overall charge of plus four on the reacting side where we need our charge on both sides of the reaction to be balanced. So we're going to cancel out this plus four charge by adding electrons on our reactant side here. So we're going to expand this and say plus four electrons. And now we need to get to that overall equation by bouncing out our electrons. So we would multiply this first half reaction by four and that would lead us to four moles of our chromium caddy on, producing four molds of our chromium three plus carry on Plus four electrons. And that is how we get to that overall reaction. When we add these two equations together, we can now cancel out these four electrons here and we'll get that overall reaction given in the prompt where we have four molds of chromium two plus kati on Plus one mole of oxygen gas plus our formals of hydride gas producing our product, we have two moles of water and sorry, that should not be a charge here. So two moles of our liquid water plus our four moles of our chromium three plus catalon. So this was our overall reaction given in the prompt. We also need to based on our half reactions written out, identify which occurs at the anodes and which occurs at the cathode. Now, if we look at our oxidation number for chromium two plus as the caddy on, it would have an oxidation state. And let's write this in red an oxidation state of plus two where on the product side it has an oxidation state of plus three being its ion charge. So we went to an increase in oxidation number because we ended up with a plus three charge on the product side here. And so because our oxidation number increases, we can say that this half reaction occurs as an oxidation, meaning that this will be at our an ode of our voltaic cell. And when we look up our standard cell potential for the oxidation of our chromium to Pluskat down in our textbooks, we would have a standard cell potential equal to negative 0.50V. And so this first half reaction is going to occur as a reduction. We also know that because we see that we gain electrons here on the react inside. And whenever we have electrons on the react inside, that means that we have a reduction reaction occurring. And so recall that our reduction reaction occurs at the cathode of our voltaic cell. And when we look up the standard cell potential for our reduction of oxygen gas in our textbooks, we find a value of 1.23V. So now we can find the standard cell potential for our overall reaction by recalling That we find in not sell the standard cell potential by taking the difference between ourself potential of our cathode, subtracted from the standard cell potential of our an ode. And so plugging in what we know from above, we have our cell potential of our cathode half reaction which we determined to be 1.23V subtracted from our cell potential of our anote half reaction being negative 0.50V. And from this difference we get an overall standard cell potential equal to 1.73V. We also want to make note of our electrons transferred. And because we canceled out four electrons from both of our reactions, we would say that N. is equal to four. Now going back to our nursed equation here, we need to figure out our E M. F. Or electric magnetic force being our self potential at nonstandard conditions. We know our standard cell potential overall. We know our electrons transferred. And we know that our reaction quotient Q is our concentration of products. Overreact ints. We're going back to the prompts. We're given our concentration for one of our Aquarius products being our concentration of our chromium three plus catalon. But we see we have a second acquis product being our concentration of H plus. And so we want to recall that we can find our concentration of hydride or H plus by using our ph so we can find 10 to the negative ph mean we would have 10 to the negative 3.25 because we're given our ph according to the prompt being 3.25. And so taking this value in our calculators will get a concentration of hydride equal to 5.6 to 3 times 10 to the negative fourth power. And we have a minimum amount of sig figs being too sick fix. So we're just going to round this down to just 5.6 times 10 to the negative fourth power as our concentration of hydride. And so now we can go into our equation where we're going to focus on that reaction quotient part. So Q. And we're going to say that Q again is equal to our concentration of our products being our concentration of H plus, multiplied by a concentration of chromium three plus and sorry correction. So we need to focus on that overall reaction here and recognize that our only product is actually our three are chromium three plus carry on here. That was our only acquis product where hydride is actually our reactant. So apologies for that mistake as well as our chromium catalon is our only other second acquis reactant. So let's fill in that acu symbol there. And so we would say that our reaction quotient is our concentration of our acquis products over our Aquarius react ints so plugging in what we know we have our concentration of our only across product being are chromium three plus, carry on. Which in our overall reaction has a coefficient of four. So that coefficient will be an exponents in the reaction quotient divided by our concentration of our acquis reactions. Where we have our formals of hydride where that coefficient of four becomes an exponents multiplied by the concentration of our second Aquarius reactant being are chromium two plus Catalan which also has a coefficient of four, which will become an exponents here. And so plugging in our values that we know in the prompt were given our concentration of our chromium three plus Catalan as 0.12 which is raised to the fourth power and then in our denominator we just found out our concentration of hydride being 5.6 times 10 to the negative fourth power raised to the fourth power, multiplied by a concentration of our chromium to Pluskat ion of 1.7 times 10 to the negative fourth power raised to the fourth power given in the prompt. And then we also have our oxygen gas which can also be included in the reaction quotient since it's in the gas phase. And so we're given its pressure in the prompt being 100.75 A. T. M. So we're going to include that as .75 and our oxygen gas does not have a coefficient. It just has a coefficient of one. So it's just going to be .75 in the denominator there. So just to be clear here, we also include our concentration of co two because it's our gaseous product And this is equal to the next step. So in our calculators we're going to carefully plug and chug this entire quotient in. And what we're going to get is our value for our reaction quotient Q equal to. And just to make a correction here, we will not include that times 10 to the negative for their for our concentration of our chromium two plus catty on. It's just given as 1.7 moller from the prompt and raise to its coefficient as an exponent being for from our overall reaction here. So sorry about that confusion. This means we can move this closer and we will get for this entire quotient a value of 33660.1 which we will write in two sig figs being our minimum number of sig figs to about 3. times 10 to the fourth power. I'm sorry. This is 1/4 power here. So now we have this value for our reaction quotient. Let's go back into our equation to calculate R E M F or cell potential and nonstandard conditions. So this term eat here. So we would say that that is equal to again, we have our cell potential under standard conditions, which we calculated here as e not sell being 1.73 volts. So we'll plug that in 1.73 volts subtracted from our quotient of 0.592 volts divided by our electrons transferred. Where we said that N is equal to four, so that's divided by four, Which is then multiplied by the log of our reaction quotient which we determined to be 3.3 times 10 To the fourth power. And so simplifying this in our next line will have 1.73V subtracted from. When we take that first quotient in parentheses, we have a value of 0.0148V. This is multiplied by when we take our log of our reaction quotient we have is 4.51851. And so now continuing our order of operations, we have 1.73V subtracted from our product, which we should get being 0.066874. This is in units of also volts. And from this difference we're going to have a value of 1.663113 as our cell potential under nonstandard conditions, a k a r e M F or electric magnetic force. And we will round this to about 366 to be one point 66V. So 1.66V is our final answer for our electric magnetic force. To complete this example. I hope everything that I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

A voltaic cell utilizes the following reaction: 4 Fe2+1aq2 + O21g2 + 4 H+1aq2 ¡ 4 Fe3+1aq2 + 2 H2O1l2 (b) What is the emf of this cell when 3Fe2+4 = 1.3 M, 3Fe3+4= 0.010 M, PO2 = 0.50 atm, and the pH of the solution in the cathode half-cell is 3.50?

Verified Solution

Key Concepts

Electrochemical Cells

Nernst Equation

pH and Hydrogen Ion Concentration