The standard reduction potentials of some half cell reactions are given below:
`PbO_(2) + 4H^(+) + 2e^(-)= Pb^(2+) + 2H_(2)O E_(0) = 1.455 V`
`MnO_(4)^(+) +8H^(+) +5e^(-) = Mn^(2+) + 4H_(2)O E_(0) = 1.51 V`
`Ce^(4+) + e^(-)hArr Ce ^(3+) E_0 = 1.61 V`
`H_2O_2 + 2H^(+) + 2e^(-) hArr 2H_(2)O E_0 = 1. 71 V`
Pick out the Incorrect statement :

Options

A. `Ce^(+4)` will oxidise `Pb^(2+)" to "PbO_2`
B. `MnO_(4)^(-)` will oxidise `Pb^(2+) " to "PbO_2`
C. `H_2O_2` will oxidise `Mn^(+2)" to " MnO_(4)^(-)`
D. `PbO_2` will oxidise `Mn^(+2)` to `MnO_(4)^(-)`

Answer

D

Explanation

Standard reduction potential of `Ce^(+4)` is greater than `PbO_2` so `Ce^(+4)` reduce itself and oxidise `Pb^(2+)`
Standard reduction potential of `MnO_4` is greater than `PbO_2` and same for `H_2O_2`
Standard reduction potential of `H_2O_2` is greater than `MnO_4` Hence it oxidises `Mn^(+2)` to `MnO^(4 -)` So option 4 is incorrect.