For the given cell `Pt_(D_(2)|D^(o+))||H^(o+)|Pt_(H_(2))`, if `E^(c-)._(D_(2)|D^(o+))=0.003V,` , what will be the ratio of `D^(o+)` and `H^(o+)` at `25^(@)C` when the reaction `D_(2) + 2H^(o+) rarr 2D^(o+)+H_(2)` attains equilibrium

Options

A. `1.34`
B. `1.24`
C. `1.124`
D. `1.45`

Answer

c

Explanation

`E_(cell)=E_(OPD_(2))=E_(RPH_(2))`
`(OP=` oxidation potential, `RP=` reduction potential `)`
`=E^(c-)._(OP_(D_(2))|D^(o+))-(0.059)/(2)log[D^(o+)]^(2)+E_(RPH^(o+)|H_(2))+0.059log[H^(o+)]^(2)`
`0=0.003-(0.059)/(2)log.([D^(o+)]^(2))/([H^(o+)]^(2))`
`([D^(o+)])/([H^(o+)])=1.124`