Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:
![Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)](https://tex.z-dn.net/?f=Pb%5E%7B2%2B%7D%28aq%29%2BZn%28s%29%5Crightarrow%20Zn%5E%7B2%2B%7D%28aq%29%2BPb%28s%29)
The half-cell reactions are:
Oxidation half reaction (anode): ![Zn\rightarrow Zn^{2+}+2e^-](https://tex.z-dn.net/?f=Zn%5Crightarrow%20Zn%5E%7B2%2B%7D%2B2e%5E-)
Reduction half reaction (cathode): ![Pb^{2+}+2e^-\rightarrow Pb](https://tex.z-dn.net/?f=Pb%5E%7B2%2B%7D%2B2e%5E-%5Crightarrow%20Pb)
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B2.303RT%7D%7BnF%7D%5Clog%20%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D%7D%7B%5BPb%5E%7B2%2B%7D%5D%7D)
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = ![25^oC=273+25=298K](https://tex.z-dn.net/?f=25%5EoC%3D273%2B25%3D298K)
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = +0.63 V
= cell potential for the reaction = ?
= 3.5 M
= ![2.0\times 10^{-4}M](https://tex.z-dn.net/?f=2.0%5Ctimes%2010%5E%7B-4%7DM)
Now put all the given values in the above equation, we get:
![E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D%28%2B0.63%29-%5Cfrac%7B2.303%5Ctimes%20%288.314%29%5Ctimes%20%28298%29%7D%7B2%5Ctimes%2096500%7D%5Clog%20%5Cfrac%7B3.5%7D%7B2.0%5Ctimes%2010%5E%7B-4%7D%7D)
![E_{cell}=0.50V](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D0.50V)
Therefore, the cell potential for this reaction is 0.50 V
Answer:
d. is the hydrostatic pressure produced on the surface of a semi-permeable membrane by osmosis.
Explanation:
Osmosis -
It is the flow of the molecules of solvent from a region of higher concentration towards the region of lower concentration via a semipermeable membrane , is known as osmosis.
Osmotic pressure -
It refers to the minimum amount of pressure , which is required to be applied to the solution in order to avoid the flow of pure solvent via the semipermeable membrane , is referred to as osmotic pressure.
Or in simple terms ,
Osmotic pressure is the pressure applied to resists the process of osmosis.
Hence ,
From the given options in the question,
The correct option regarding osmotic pressure is d.
A science conclusion should simply have a statement identifying whether your experiment’s aim turned out to be true or false
Answer:
The correct option is: (D) -2.4 kJ/mol
Explanation:
<u>Chemical reaction involved</u>: 2PG ↔ PEP
Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol
Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)
Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)
Reactant concentration: 2PG = 0.5 mM
Product concentration: PEP = 0.1 mM
Reaction quotient: ![Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2](https://tex.z-dn.net/?f=Q_%7Br%7D%20%3D%5Cfrac%7B%5Cleft%20%5B%20PEP%20%5Cright%20%5D%7D%7B%5Cleft%20%5B%202PG%20%5Cright%20%5D%7D%20%3D%20%5Cfrac%7B0.1%20mM%7D%7B0.5%20mM%7D%20%3D%200.2)
<u>To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:</u>
![\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%20%5CDelta%20G%5E%7B%5Ccirc%20%7D%20%2B%202.303%20R%20T%20log%20Q_%7Br%7D)
![\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%201.7%20kJ%2Fmol%20%2B%20%5B2.303%20%5Ctimes%20%288.314%20%5Ctimes%2010%5E%7B-3%7D%20kJ%2F%28K.mol%29%29%5Ctimes%20%28310.15%20K%29%5D%20log%20%280.2%29)
![\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%201.7%20%2B%20%5B5.938%5D%20%5Ctimes%20%28-0.699%29%20%3D%201.7%20-%204.15%20%3D%20%28-2.45%20kJ%2Fmol%29)
<u>Therefore, the Gibb's free energy change at 37° C (310.15 K): </u><u>ΔG = (-2.45 kJ/mol)</u>