MnO_(4)^(-)+Fe^(2+)longrightarrow Mn^(2+)+Fe^(3+)

MnO_(4)^(-) + 5Fe^(2+) + 8H^(+) rightarrow Mn^(2+) + 5Fe^(3=) + 4H_(2)O E^(@) = 0.743V What is the value of E for the cell based on the reaction above at 25^(@)C for the following conditons? {:(MnO_(4)^(-),Fe^(2+)),(2.3 xx 10^(-2), 1 xx 10^(-4)M),([Mn^(2+)],Fe^(3+)):}

Balance the following equations by oxidation no. method. MnO_4^(-) + Fe^(2+) rarr Mn^(2+) + Fe^(3+) (Acidic medium) .

Balance the equation by ion-electron method : MnO_(4)^(-) + Fe^(2+) to Mn^(2+) + Fe^(3+) + H_(2)O (in acid medium)

Assertion: In the reaction, MnO_(4)^(-)+5Fe^(2+)+8H+ rarr Mn^(2+)+5Fe^(3+)+4H_(2)O , MnO_(4)^(-) acts as oxidising agent. Reason: In the above reaction, n - factor is 5 .

KMnO_(4) reacts with ferrous ammonium sulphate according to the equation MnO_(4)^(-)+5Fe^(2+)+8H^(+) rarr Mn^(2+)+5Fe^(3+)+4H_(2)O , here 10 ml of 0.1 M KMnO_(4) is equivalent to

Balance the following equation by oxidation number method. MnO_(4)^(-)+Fe^(2+)rarrMn^(2+)+Fe^(3+) (acidic medium)

Balance the following reaction by the oxidation number method - MnO_(4)^(-)+Fe^(+2)rarrMn^(+2)+Fe^(+3)

Balance the following reaction by the oxidation number method - MnO_(4)^(-)+Fe^(+2)rarrMn^(+2)+Fe^(+3)