Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
The change is that the piston gets hotter? (Theres more heat in it)
This problem is providing the basic dissociation constant of ibuprofen (IB) as 5.20, its pH as 8.20 and is requiring the equilibrium concentration of the aforementioned drug by giving the chemical equation at equilibrium it takes place. The obtained result turned out to be D) 4.0 × 10−7 M, according to the following work:
First of all, we set up an equilibrium expression for the given chemical equation at equilibrium, in which water is omitted for it is liquid and just aqueous species are allowed to be included:
![Kb=\frac{[IBH^+][OH^-]}{[IB]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BIBH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BIB%5D%7D)
Next, we calculate the concentration of hydroxide ions and the Kb due to the fact that both the pH and pKb were given:

![[OH^-]=10^{-5.8}=1.585x10^{-6}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-5.8%7D%3D1.585x10%5E%7B-6%7DM)

Then, since the concentration of these ions equal that of the conjugated acid of the ibuprofen (IBH⁺), we can plug in these and the Kb to obtain:
![6.31x10^{-6}=\frac{(1.585x10^{-6})(1.585x10^{-6})}{[IB]}](https://tex.z-dn.net/?f=6.31x10%5E%7B-6%7D%3D%5Cfrac%7B%281.585x10%5E%7B-6%7D%29%281.585x10%5E%7B-6%7D%29%7D%7B%5BIB%5D%7D)
Finally, we solve for the equilibrium concentration of ibuprofen:
![[IB]=\frac{(1.585x10^{-6})(1.585x10^{-6})}{6.31x10^{-6}}=4.0x10^{-7}](https://tex.z-dn.net/?f=%5BIB%5D%3D%5Cfrac%7B%281.585x10%5E%7B-6%7D%29%281.585x10%5E%7B-6%7D%29%7D%7B6.31x10%5E%7B-6%7D%7D%3D4.0x10%5E%7B-7%7D)
Learn more:
(Weak base equilibrium calculation) brainly.com/question/9426156