<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.
<u>Explanation:</u>
We are given:
The substance having highest positive 
potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.
Aluminium will undergo oxidation reaction and will get oxidized.
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the 
of the reaction, we use the equation:
Hence, the standard electrode potential of the cell is 4.53 V.
Even though two grams seemed to disappear or vanish, the law of conversation of mass still holding true. Mercuric oxide, when heated, forms a gas of mercury and oxygen. During the investigation, some gas could have escaped or evaporated.
Answer:
It is basically a way of telling you how to solve for different variables in the equation d=m/v
Explanation:
Explanation:
Given -
- An organic compound gives H₂ gas with Na
- On treatment with alkaline iodine it gives yellow ppt.
- On oxidation with CrO₃/H⁺ forms an aldehyde (C₂H₄O)
To Find -
- Name the compound and write the reaction involved
Now,
Let A be the organic compound.
Then,
- A + Na → + H₂↑
- A + I₂ → CHI₃ (yellow ppt.)
- A + CrO₃ + H⁺ → C₂H₄O
Now,
Here we see that compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives aldehyde.
- Functional group of aldehyde = —CHO
And It forms only 2 Carbon aldehyde it means, It is Ethanal (CH₃CHO).
Compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives ethanal.
It means,
We know that 1° alcohol on oxidation gives aldehyde.
Here it gives 2 Carbon aldehyde.
It means,
Here 2 Carbon and 1° alcohol is used.
Now,
Its cleared that Compound A is Ethanol.
Reaction Involved -
- CH₃CH₂OH + Na → CH₃CH₂O⁻Na⁺ + H₂↑
- CH₃CH₂OH + I₂ + OH⁻ → CHI₃↓ + HCOO⁻ + HI + H₂O
- CH₃CH₂OH + CrO₃ + H⁺ → CH₃CHO
