How much will the reduction potential of a hydrogen electrode change when its solution initially at pH=0 is neutralized to pH = 7?

This is the relation between emf of cell and pH. Emf of hydrogen is represented by formula:
${\mathrm{E}}_{{\mathrm{H}}_2}^0=\frac{-2.303\mathrm{RT}}{\mathrm{F}}\times \mathrm{pH}$   

where, R and F are constant .
If we consider emf of hydrogen at ${25}^0$C then
${\mathrm{E}}_{{\mathrm{H}}_2}^0(\mathrm{at} {25}^0\mathrm{c})=(-0.0591)\times \mathrm{pH}$
Now, for pH=0 by putting values
${\mathrm{E}}_{{\mathrm{H}}_2}^0=0$
For pH=7 by putting values
${\mathrm{E}}_{{\mathrm{H}}_2}^0=-0.0591\times 7=-0.41$, it means emf of hydrogen decreases by 0.41 V.

Hence option D is correct.