निम्नलिखित सेल में एनोडीय विलयन का pH मान है :
`Pt, H_(2)(1"atm") "|" H^(+)(xM)"||" H^(+) (1M)"|" H_(2) (1"atm"), Pt`
यदि `E_("cell")=0.2364` V दिया है।

Answer

4

Explanation

कैथोड परः
`2H^(+) (1M) +2e^(-) to H_(2) `(गैस)
ऐनोड पर :
`H_(2)(g) to 2H^(+) (xM) +2e^(-)`
अब:
`E_("Cell")=E_("Cell")^(@) -(0.0591)/(2)log ((([H^(+)]_("ऐनोड"))^(2))/(([H^(+)]_("कैथोड"))^(2)))`
`E_("Cell") =E_("Cell")^(@) -(0.0591)/(2)log ((([H^(+)]_("ऐनोड"))^(2))/(([H^(+)]_("कैथोड"))^(2)))^(2)`
`E_("Cell")=E_("Cell")^(@) -(0.0591)/(2)log ((x)/(1))^(2)`
`0.2364= 0 -(0.0591)/(2) log((x)/(1))^(2)`
`0.2364 xx 2 = -0.0591 log ((x)/(1))^(2)`
`(0.2364 xx 2)/(0.0591) = -log ((x)/(1))^(2)`
`-log ((x)/(1))^(2) = (0.2364 xx 2)/(0.0591)=8`
`-2 log ((x)/(1))=8`
`log ((x)/(1)) = -4 `
`x=10^(-4)`
`pH= -log (x)`
`pH=4`