Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. give below are a set of half-cell reaction (acidic medium) along with their `E^(@)` (V with respect to normal hydrogen electrode) values. using this data obtain the correct explanation to folloing Questions. `I_(2)+2e^(-)to2I^(-)" "E^(0)=0.54` `Cl_(2)+2e^(-)to2Cl^(-)" "E^(0)=1.36` `Mn^(3+)+e^(-)toMn^(2+)" "E^(0)=1.50` `Fe^(3+)+e^(-)toFe^(2+)" "E^(0)=0.77` `O_(2)+4H^(+)+4e^(-)to2H_(2)O" "E^(0)=1.23` Q. Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and `H_(2)SO_(4)` in presence of air gives a prussian blue precipitate. the blue colour is due to the formation ofA. `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 reaction play a pivotal role in chemistry and biology. The value

1.	Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. give below are a set of half-cell reaction (acidic medium) along with their `E^(@)` (V with respect to normal hydrogen electrode) values. using this data obtain the correct explanation to folloing Questions. `I_(2)+2e^(-)to2I^(-)" "E^(0)=0.54` `Cl_(2)+2e^(-)to2Cl^(-)" "E^(0)=1.36` `Mn^(3+)+e^(-)toMn^(2+)" "E^(0)=1.50` `Fe^(3+)+e^(-)toFe^(2+)" "E^(0)=0.77` `O_(2)+4H^(+)+4e^(-)to2H_(2)O" "E^(0)=1.23` Q. Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and `H_(2)SO_(4)` in presence of air gives a prussian blue precipitate. the blue colour is due to the formation ofA. `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 `Na+C+Nto underset("Sodium extract")(NaCN)` `Fe^(2+)+6CN^(-)to[Fe(CN)_(6)]^(4-)` `4Fe^(3+)+3[Fe(CN)_(6)]^(4-)to underset("Prussian blue ppt.")(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^(-)`
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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