Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
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
Answer:
It starts to become a red giant!
Explanation:
