The cell emf in a cell with cell reaction Cr_2O_7^(2-) + 14H^(+) + 6Fe^(+2) to 2Cr^(+3) + 6Fe^(+3) + 7H_2O could be increased above the standard emf by
Calculate the potential of the half-cell containing 0.1 M K_(2)Cr_(2)O_(7(aq)), 0.2M Cr_((aq))^(3+)and 1.0xx10^(-4) M H_((aq))^(+). The half-reaction Cr_(2) O_(7(aq))^(2-)+ 14 H_((aq))^(+)+6e^(-)to 2 Cr_((aq))^(3+)+7H_(2)O_((l)) (E^(0) of Cr_(2)O_(7)^(2-) //Cr^(3+)=1.33V)
The standard electrode potential for the following reaction is + 1.33V.What is potential at pH=2.0? Cr_2O_7^(2-) (aq. 1M) + 14H^(+)(aq) + 6e^(-) to 2Cr^(3+) (aq. 1M0 + 7H_2O(l)
For the reaction 5 Br(aq) + BrO_(3)^(-)(aq) + 6H^(+)(aq) to 3 Br_(3)(aq) + 3H_(2)O(l) If, -(Delta[Br])/(Deltat) = 0.05 mol L ^(-1)mi n^(-1) , -(Delta[BrO_(3)])/(Deltat) in mol L ^(-1)mi n^(-1) is
Balance the following redox reactions by ion-electron method : H_(2)O_(2)(aq)+Fe^(2+)(aq)to Fe^(3+)(aq)+H_(2)O(I) (in acidic solution )
Consider the reactions 2S_(2)O_(3)^(2-)(aq)+I_(2)(s)toS_(4)O_(6)^(2-)(aq)+2I^(-)(aq) S_(2)O_(3)^(2-)(aq)+2Br_(2)(l)+5H_(2)O(l)to2SO_(4)^(2-)(aq)+4Br^(-)(aq)+10H^(+)(aq) Why does the same reductant, thiosulphate react differently with iodine and bromine ?