Answer: The value of
for the half-cell reaction is 0.222 V.
Explanation:
Equation for solubility equilibrium is as follows.
![AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)](https://tex.z-dn.net/?f=AgCl%28s%29%20%5Crightleftharpoons%20Ag%5E%7B%2B%7D%28aq%29%20%2B%20Cl%5E%7B-%7D%28aq%29)
Its solubility product will be as follows.
![K_{sp} = [Ag^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)
Cell reaction for this equation is as follows.
![Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)](https://tex.z-dn.net/?f=Ag%28s%29%7C%20AgCl%28s%29%7CCl%5E%7B-%7D%280.1%20M%29%7C%7C%20Ag%5E%7B%2B%7D%281.0%20M%29%7C%20Ag%28s%29)
Reduction half-reaction:
, ![E^{o}_{Ag^{+}/Ag} = 0.799 V](https://tex.z-dn.net/?f=E%5E%7Bo%7D_%7BAg%5E%7B%2B%7D%2FAg%7D%20%3D%200.799%20V)
Oxidation half-reaction:
,
= ?
Cell reaction: ![Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)](https://tex.z-dn.net/?f=Ag%5E%7B%2B%7D%28aq%29%20%2B%20Cl%5E%7B-%7D%28aq%29%20%5Crightarrow%20AgCl%28s%29)
So, for this cell reaction the number of moles of electrons transferred are n = 1.
Solubility product, ![K_{sp} = [Ag^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)
= ![1.77 \times 10^{-10}](https://tex.z-dn.net/?f=1.77%20%5Ctimes%2010%5E%7B-10%7D)
Therefore, according to the Nernst equation
At equilibrium,
= 0.00 V
Putting the given values into the above formula as follows.
![E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}](https://tex.z-dn.net/?f=E%5E%7Bo%7D_%7Bcell%7D%20%3D%20%5Cfrac%7B0.0592%7D%7B1%7D%20log%20%5Cfrac%7B1%7D%7BK_%7Bsp%7D%7D)
= ![0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}](https://tex.z-dn.net/?f=0.0591%20V%20%5Ctimes%20log%20%5Cfrac%7B1%7D%7B1.77%20%5Ctimes%2010%5E%7B-10%7D%7D)
= 0.577 V
Hence, we will calculate the standard cell potential as follows.
![E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}](https://tex.z-dn.net/?f=E%5E%7Bo%7D_%7Bcell%7D%20%3D%20E%5E%7Bo%7D_%7Bcathode%7D%20-%20E%5E%7Bo%7D_%7Banode%7D)
![0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}](https://tex.z-dn.net/?f=0.577%20V%20%3D%20E%5E%7Bo%7D_%7BAg%5E%7B%2B%7D%2FAg%7D%20-%20E%5E%7Bo%7D_%7BAgCl%2FAg%7D)
![0.577 V = 0.799 V - E^{o}_{AgCl/Ag}](https://tex.z-dn.net/?f=0.577%20V%20%3D%200.799%20V%20-%20E%5E%7Bo%7D_%7BAgCl%2FAg%7D)
= 0.222 V
Thus, we can conclude that value of
for the half-cell reaction is 0.222 V.
Answer;
-H2O, water
Explanation;
-Hydrocarbons undergoes combustion to form water and carbon dioxide. Propane (C3H8) is a hydrocarbon that belongs to the homologous series alkanes. Thus, propane will undergo combustion to form 3 molecules of carbon dioxide and 4 molecules of water.
Thus the complete equation will be;
-C₃H₈ + 5O₂ = 3CO₂ + 4H₂O
Answer:
The solution is given below
Explanation:
Heat, q= mc∆T
q= 125g x 4.18 J/g∙°C x (21.18x- 24.28) °C
q= -1619.75J
NEGATIVE SIGN INDICATES THAT HEAT IS ABSORBED.
Enthalpy Change, ∆H = 1619.75 7/ 10.5 g
= 154.26 J/g
No. of moles of KBr = Mass of KBr/ Molecular Weight of KBr
=10.5g/119gmol-1
=0.088 mol
∆H= 1619.75 J/ 0.088 mol
= 18.41 kJ/mol
Answer:
He took a deep breath and splashed some water on his face.
Explanation:
I took the test on Edmentum.
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