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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 equilibriumA. `1.34`B. `1.24`C. `1.124`D. `1.45` |
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Answer» Correct Answer - c `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` |
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