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Redox reactions play a pivotal role in c...

Redox reactions play a pivotal role in chemistry and biology. The values standard redox potential `(E^(o))` of two half cell reactions decided which way the reaction is expected to preceed. A simple example is a Daniell cell in which zinc goes into solution and copper sets deposited. Given below are a set of half cell reactions `(` acidic medium `)` along with their `E^(c-)(V` with respect to normal hydrogen electrode `)` values. Using this data, obtain correct explanations for Question.
`I_(2)+2e^(-) rarr 2I^(c-)," "E^(c-)=0.54`
`Cl_(2)+2e^(-) rarr 2Cl^(c-), " "E^(c-)=1.36`
`Mn^(3+)+e^(-) rarr Mn^(2+), " "E^(c-)=1.50`
`Fe^(3+)+e^(-) rarr Fe^(2+)," "E^(c-)=0.77`
`O_(2)+4H^(o+)+4e^(-) rarr 2H_(2)O, " "E^(c-)=1.23`
Sodium fusion extract obtained from aniline on treatment with iron `(II)` sulphate and `H_(2)SO_(4)` in the presence of air gives a Prussion blue precipitate. The blue colour is due to the formation of

A

`Fe_(4)[Fe(CN)_(6)]_(3)`

B

`Fe_(3)[Fe(CN)_(6)]_(2)`

C

`Fe_(4)[Fe(CN)_(6)]_(2)`

D

`Fe_(3)[Fe(CN)_(6)]_(3)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem regarding the formation of a Prussian blue precipitate from the sodium fusion extract obtained from aniline, we will follow these steps: ### Step-by-Step Solution: 1. **Understanding the Sodium Fusion Extract**: - The sodium fusion extract from aniline contains sodium cyanide (NaCN). This is due to the presence of carbon and nitrogen in aniline, which reacts with sodium. 2. **Reaction with Iron(II) Sulfate**: - When the sodium cyanide is treated with iron(II) sulfate (FeSO4), the cyanide ion (CN⁻) from NaCN reacts with Fe²⁺ ions. This reaction leads to the formation of a complex ion: \[ \text{Fe}^{2+} + \text{CN}^- \rightarrow \text{[Fe(CN)}_6]^{4-} \] - Here, the complex formed is [Fe(CN)₆]⁴⁻. 3. **Oxidation in Acidic Medium**: - When this complex is treated with sulfuric acid (H₂SO₄), the Fe²⁺ ions in the complex can be oxidized to Fe³⁺ ions: \[ \text{Fe}^{2+} \xrightarrow{H_2SO_4} \text{Fe}^{3+} \] 4. **Formation of Prussian Blue**: - The Fe³⁺ ions then react with the cyanide ions to form Prussian blue, which is represented by the formula: \[ \text{Fe}_4[\text{Fe(CN)}_6]_3 \] - This complex is known for its blue color, which is why the precipitate appears blue. 5. **Conclusion**: - Therefore, the blue color of the precipitate obtained is due to the formation of the Prussian blue complex, \(\text{Fe}_4[\text{Fe(CN)}_6]_3\). ### Final Answer: The blue color is due to the formation of Prussian blue, \(\text{Fe}_4[\text{Fe(CN)}_6]_3\). ---
To solve the problem regarding the formation of a Prussian blue precipitate from the sodium fusion extract obtained from aniline, we will follow these steps: ### Step-by-Step Solution: 1. **Understanding the Sodium Fusion Extract**: - The sodium fusion extract from aniline contains sodium cyanide (NaCN). This is due to the presence of carbon and nitrogen in aniline, which reacts with sodium. 2. **Reaction with Iron(II) Sulfate**: ...
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Redox reactions play a pivotal role in chemistry and biology. The values standard redox potential (E^(c-)) of two half cell reactions decided which way the reaction is expected to preceed. A simple example is a Daniell cell in which zinc goes into solution and copper sets deposited. Given below are a set of half cell reactions ( acidic medium ) along with their E^(c-)(V with respect to normal hydrogen electrode ) values. Using this data, obtain correct explanations for Question. I_(2)+2e^(-) rarr 2I^(c-)," "E^(c-)=0.54 Cl_(2)+2e^(-) rarr 2Cl^(c-), " "E^(c-)=1.36 Mn^(3+)+e^(-) rarr Mn^(2+), " "E^(c-)=1.50 Fe^(3+)+e^(-) rarr Fe^(2+)," "E^(c-)=0.77 O_(2)+4H^(o+)+4e^(-) rarr 2H_(2)O, " "E^(c-)=1.23 While Fe^(3+) is stable, Mn^(3+) is not stable in acid solution because

Redox reactions play a pivotal role in chemistry and biology. The values standard redox potential (E^(c-)) of two half cell reactions decided which way the reaction is expected to preceed. A simple example is a Daniell cell in which zinc goes into solution and copper sets deposited. Given below are a set of half cell reactions ( acidic medium ) along with their E^(c-)(V with respect to normal hydrogen electrode ) values. Using this data, obtain correct explanations for Question. I_(2)+2e^(-) rarr 2I^(c-)," "E^(c-)=0.54 Cl_(2)+2e^(-) rarr 2Cl^(c-), " "E^(c-)=1.36 Mn^(3+)+e^(-) rarr Mn^(2+), " "E^(c-)=1.50 Fe^(3+)+e^(-) rarr Fe^(2+)," "E^(c-)=0.77 O_(2)+4H^(o+)+4e^(-) rarr 2H_(2)O, " "E^(c-)=1.23 Among the following, identify the correct statement. (a)Chloride ion is oxidized by O_(2) . (b) Fe^(2+) is oxidized by iodine. (c)Iodide ion is oxidized by chlorine (d) Mn^(2+) is oxidized by chlorine.

Given below are a set of half-cell reactions (acidic medium) along with their E_(@) with respect to normal hydrogen electrode values. Using the data obtain the correct explanation to question given below. {:(I_(2)+2e^(-)rarr2I^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-),E^(@)=1.36),(Mn^(2+)+e^(-)rarrMn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):} Among the following, identify the correct statement:

Given below are a set of half-cell reactions (acidic medium) along with their E_(@) with respect to normal hydrogen electrode values. Using the data obtain the correct explanation to question given below. {:(I_(2)+2e^(-)rarr2I^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-),E^(@)=1.36),(Mn^(2+)+e^(-)rarrMn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):} While Fe^(2+) is stable, Mn^(3+) is not stable in acid solution because:

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