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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. Among the following, identify the correct statementA. Chloride ion oxidised `O_(2)`B. `Fe^(2+)` is oxidised by iodineC. Iodide ion is oxidised by chlorineD. `Mn^(2+)` is oxidised by chlorine |
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Answer» Correct Answer - C `2I^(-)+Cl_(2)toI_(2)+2Cl^(-)` `E^(o)=E_(I^(-)//I_(2))^(o)+E_(Cl_(2)//Cl^(-))^(o)` `=-0.54+1.36` `E^(o)=0.82V` `E^o` is positive hence iodide ion is oxidized by chlorine. |
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