Redox reactions play a pivotal role in chemistry and biology. The value of standard reduction potetials `(E^(@))` of the two half cells reactions decide which way the reaction is expected to proceed. A simple example is of Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with with `E^(@)` values. Using the data obtain the correct explanation to the questions that are mentioned. `I_(2)+e^(-) to 2I^(-) , E^(@)=0.54" V "` `Cl_(2)+2e^(-) to 2Cl^(-) , E^(@)=1.36" V"` `Mn^(2+)+2e^(-) to Mn , E^(@)=1.50" V"` `Fe^(3+)+e^(-) to Fe^(2+) , E^(@)=0.77" V"` `O_(2)+4H^(+)+4e^(-) to 2H_(2)O , E^(@)=1.23" V"`. Sodium fusion extract obtained from aniline, on treatment with iron (II) sulphate and `H_(2)SO_(4)` in the presence of air, gives a prussian 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)`.

Redox reactions play a pivotal role in chemistry and biology. The valu

1.	Redox reactions play a pivotal role in chemistry and biology. The value of standard reduction potetials `(E^(@))` of the two half cells reactions decide which way the reaction is expected to proceed. A simple example is of Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with with `E^(@)` values. Using the data obtain the correct explanation to the questions that are mentioned. `I_(2)+e^(-) to 2I^(-) , E^(@)=0.54" V "` `Cl_(2)+2e^(-) to 2Cl^(-) , E^(@)=1.36" V"` `Mn^(2+)+2e^(-) to Mn , E^(@)=1.50" V"` `Fe^(3+)+e^(-) to Fe^(2+) , E^(@)=0.77" V"` `O_(2)+4H^(+)+4e^(-) to 2H_(2)O , E^(@)=1.23" V"`. Sodium fusion extract obtained from aniline, on treatment with iron (II) sulphate and `H_(2)SO_(4)` in the presence of air, gives a prussian 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)`.
Answer» Correct Answer - A (a) `Na +C+N to NaCN` `Fe^(2+)+6CN^(-) to [Fe(CN)_(6)]^(4-)` `4Fe^(3+)+3[Fe(CN)_(6)]^(4-) to underset("Prussian blue")(Fe_(4)[Fe(CN)_(6)]_(3))`

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Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of a solution for a weak and a strong electrolyte.
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Based on the data given below, the correcy order of reducing power is: `Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V` `Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V` `Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V`A. `Br^(-) lt Fe^(2+) lt Al`B. `Fe^(2+) lt Al lt Br^-`C. `Al lt Br^(-) lt Fe^(2+)`D. `Al lt Fe^(2+) lt Br^(-)`
What are elementary reactions ?

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