1.

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

Answer»

Increases by 0.059 V
Decreases by 0.59 V
Increases by 0.41 V
Decreases by 0.41 V

Solution :`H^(+)+e^(-)to(1)/(2)H_(2)`
`E_(RED)=E_(red)^(@)-(0.0591)/(2)"log"(1)/([H^(+)])`
`=E_(red)^(@)+0.0591log[H^(+)]`
`E_(red)^(@)=0`. When `pH=0,[H^(+)=]10^(@)=1M`
`thereforeE_(red)=0`
when pH=7,`[H^(+)]=10^(-7)M`
`E_(red)=0.0591" log "10^(-7)(0.0591)(-7)`
`=-0.4137`
THUS, `E_(red)` decreases by 0.41 V.


Discussion

No Comment Found

Related InterviewSolutions