An electrochemical cell is based on these two half-reactions: Ox: Pb(s) -> Pb2+ (aq, 0.10 M) + 2 e- Red: MnO4-(aq, 1.50 M) + 4 H+(aq, 2.0 M) + 3 e- -> MnO2(s) + 2 H2O(l) Calculate the cell potential at 25 °C.

Video transcript

Hey everyone in this example, we need to consider the following half reactions and an electrochemical cell and determine the cell potential at 25°C. Our first step is to write out our two half reactions. So for our first half reaction were given an oxidation where we have solid iron to form the iron, three plus catalon and three electrons are added on. And then for our reduction half reaction we have solid lead oxide plus four hydrogen atoms plus sulfate and ion and two electrons produces lead, sulfate, solid and two moles of liquid water. Now, all of our atoms in both of these half reactions are balanced. However, we need to make sure that the electrons are also balanced and right now we have three electrons in the first half reaction and just two in the second half reaction. So we're going to take both of these reactions and multiply them by a factor so that we have the same amount of electrons for both half reactions. So we're gonna multiply the first half reaction by two and then we're gonna multiply the second half reaction by three. And so this would create two new equations where below we will now have two moles of solid iron producing two moles of iron, three plus cat ion plus six electrons now and then for our second reaction where we have our reduction, we have three moles of lead oxide plus 12 moles of hydride plus three moles of the sulfate an ion and then plus six electrons produces on our product side, we have three moles of lead sulfate plus six moles of liquid water. Now that we have the same amount of electrons for each of our reactions. We can go ahead and cancel the electrons out because we have it on the product side here and on the reactive side in our reduction reaction, we're going to add these two reactions together so that we have just one equation and what we're going to get is two moles of solid iron plus three moles of lead oxide Plus 12 moles of hydride, Plus three moles of the sulfate an ion. And sorry, that's two produces on our product side, two moles of iron three plus cat ion plus three moles of lead sulfate and then plus six moles of water to complete our reaction here. Another important mention is noting. Based on our standard electrode or standard reduction potential tables in our textbooks or online, we're going to look up for the oxidation of solid iron. We would have a self potential value equal to negative 0.36 volts. Also looking up for the reduction of solid lead oxide. We would have a self potential value and sorry, that should be equal to Positive 1.69V. And we can see that because this value is positive. This would occur at our cathode, whereas our first value here is going to occur at are an ode. Our next step is to calculate our standard cell potential for reaction, which we should recall is calculated by taking the standard cell potential and setting that equal to the cell potential of our cathode. Subtracted from the cell potential of our an ode. And so what we would have is our self potential of our cathode, which above we stated is 1.69V subtracted from the cell potential of our an ode which we stated above is negative. 0.036V. This is going to give us our standard cell potential equal 1.7-6V. So we need to calculate the cell potential for a reaction at 25°C. Right now we only know our standard cell potential. And what we're going to do to find the cell potential of the reaction is recall are nursed equation where we take the cell potential e cell and set that equal to our standard cell potential. Subtracted from our new equation constant 0.0592V divided by N. Which is then multiplied by the log of Q. Our ratio between our ions between our anodes and cathodes. So recall that Q is equal to the concentration of products divided by our concentration of reactant. And this is in reference to our ions. So we would say that for our concentration of ion products we have just iron three plus and this has a coefficient of two. So we're going to raise this to the second power as we see here. Then in our denominator we're going to have a concentration of our ion reactant where we have hydride as an ion which is going to be raised to an exponent of 12. And this is multiplied by a concentration of our other ion on our reactant side which is our sulfate an ion. So S. 042 minus which is raised to a power of three as we see here in our equation. So we're gonna go ahead and calculate our self potential and we would say that this is equal to our standard cell potential which above the stated is 1.7 to 6 volts which is then subtracted from our nerds equation constant 0.592 volts divided by n where n we should recognize as our electrons transferred. So above we would see that we canceled out six electrons in both of our half reactions. So six electrons are transferred. So we would plug in six in our denominator. And then this is multiplied by our log of Q. Where we're going to fill in this part here for our concentration. So in our numerator we have for our concentration of iron three plus, given in the prompt a value of 0.200 squared. And this is in units of polarity. And then in our denominator for our concentration of hydride. The prompt tells us that we have a value of 2.10 moller. This is to the 12th power, multiplied by a concentration of sulfate an ion Which has a value of 1.60 Moller raised to the third power. So we're going to plug that down below for Q. So now simplifying things, what we should get is that our Self potential is equal to a value of 1.7-6V subtracted from 9.8667 times 10 to the negative third power volts which is then multiplied by the log of our quotient here Which should give us a value equal to negative 5.8769. So simplifying this further, we would get that our self potential is equal 1.7-6V added to the product of these two values here, which in our calculators a value of zero and sorry that should be a pen. So zero 058. And when we take the sum here we're going to get our self potential value equal to 1.78 volts. So this here would complete this example as our final answer for our self potential of our reaction at 25 degrees Celsius. So I hope that everything I read was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video