A student titrates a cathode cell of a zinc-lead galvanic cell with potassium dichromate under acidic conditions. Lead dication forms an insoluble precipitate with dichromate dianion. As lead concentration decreases, cell potential decreases.
(Pb+2) + Zn -> (Zn+2) + Pb
Cr2O7 + Zn -> (Zn+2) + (Cr+3)
*Equations not balanced*
Reduction voltage of lead, zinc, and chromate is -.13, -.76, and 1.33
Question: Observed voltage reaches minimum when....
1. Initial lead concentration > added chromate concentration
2. Initial lead concentration = added chromate concentration
3. Moles of initial Pb = Moles of initial chromate added
4. Moles of initial Pb > Moles of initial chromate added
Answer: #3, Observed voltage is lowest when lead and chromate are closest to zero.
The book also says that excess chromate is reduced after all the lead has been reduced. But wouldn't chromate have priority since its reduction half reaction has the largest positive potential?
In order to approach this question, I would use physics and a little general chemistry.
V=kq/r
In this case, our potential difference is created by separating a positive charge and a negative charge, thus creating an E field, and when connecting the two terminals with a conductor, electrons will flow from the negative to the positive terminal, opposite the direction of the current.
The prompt states that the student is titrating a cathode cell of a zinc-lead galvanic cell with potassium dichromate under acidic conditions. Well, if we know that this is a galvanic cell, then in order for the reaction to be spontaneous, the potential of the entire cell must be positive. AnOX RedCAT. Oxidation occurs at the anode and reduction occurs at the cathode. Since they are titrating the cathode cell with potassium dichromate under acidic conditions, they must be adding potassium dichromate to the side in which reduction is occurring (cathode).
If the cell is galvanic, the deltaV must be positive. The reduction potential for lead is -0.13. That for zinc is -0.76. By the general formula, Cell potential= Eoxidation + Ereduction.
Then for the cell potential to be positive, zinc must be oxidized and lead must be reduced, because the only way to have a positive number when combining -0.13 and -0.76 is if -0.76 is positive. Reduction occurs at the cathode, so at the cathode the following reaction takes place: Pb^(2+) + 2e- ---> Pb(s)
At the anode: Zn(s)--> Zn^(2+) +2e-
So now we know that electrons are flowing towards the side where solid lead is forming. This makes sense because the problem says that as lead concentration decreases, cell potential will decrease. As the concentration of lead cation decreases, formation of solid lead slows and the battery dies.
Now, the student is titrating the cathode with potassium dichromate under acidic conditions. Lead dication forms an insoluble precipitate with dichromate dianion. Because these two are being combined in the same container, then some other reactions can take place:
reaction 1: K2Cr2O7 ---> 2K^(+) + Cr2O7^(2-)
reaction 2: Cr2O7^(2-) + Pb^(2+)---> PbCr2O7
In the above reaction, Pb combines with the dichromate dianion, as given in the prompt. As previously mentioned, AS LEAD CONCENTRATION DECREASES, CELL POTENTIAL DECREASES. So adding the potassium dichromate will decrease the lead concentration because the potassium dichromate will dissociate and the dichromate dianion will combine with lead to form an insoluble precipitate. And because ALL of this occurs in one side of the battery, then the amount of Pb^(2+) available to be reduced by electrons flowing from the anode is decreased, which leads to a decrease in cell potential.
Now, the question wants to know when does the observed voltage reach a minimum. Well, this will occur when lead available for reduction by movement of charge from anode to cathode is at a minimum. So, as the lead concentration decreases, so will voltage of the cell. It must be noted before answering the question that if potassium dichromate is added to solution, then potassium and dichromate dissociate. Which means that adding the potassium dichromate will cause yet another reaction to take place, which you listed:
K2Cr2O7 ---> 2K^(+) + Cr2O7 ^(2-)
The reduction potential of chromate is 1.33. And we know that the more positive the reduction potential, the greater a species affinity for electrons and desire for reduction. In Cr2O7^(2-) chromium has an oxidation state of plus six ( 7(-2) +2x = -2) Well, when it is present in the cathode, electrons are flowing towards the dichromate anion, which can reduce it, yielding a deltaV for this reduction of 1.33.
So you can look at it like you have two batteries instead of one. The reduction of chromate and oxidation of zinc will yield a deltaV of = 1.33 +0.76=2.09
And that of the reduction of lead and oxidation of zinc will yield a delta V of = 0.76-0.13= 0.63.
So even when all the lead is used up, excess chromate will still make the deltaV of the cell pretty large. But when lead is used up, only excess chromate will be reduced. When excess chromate reduction goes to completion, the battery will die and deltaV cell will reach a minimum. And we know that 1 mol Pb^(2+) combines with 1 mol Cr2O7^(2-) to make one mol of PbCr2O7 (s).
So, if the initial moles of Pb equal the initial moles of chromate added, then Pb can combine with dichromate anion and form the precipitate, which will result in the concentrations of both Pb^2+ and dichromate anion going to zero essentially and really no excess of either reactant will be around to be reduced. Once they go to zero, the battery will die because there is no oxidizing agent left in the cathode. Answer choice two is wrong because it does not tell you about the number of moles that are actually being combined together. You can add 2 ml of a 3M solution of chromate to an initially 3M solution of lead (the added concentration will still be 3M chromate) but in the combined solution, you will have far more lead than chromate.
I hope this helps answer your question!
