Determine at `298` for cell:
`Pt|Q, QH_(2), H^(+) ||1M KCI |Hg_(2)CI_(2)|Hg(l)|Pt`
(a) It's emf when `pH = 5.0`
(b) the `pH` when `E_(cell) = 0`
(c) the positive electrode when `pH = 7.5`
given `E_(RP(RHS))^(@) = 0.28, E_(RP(LHS))^(@) = 0.699`

Answer

(a) `-0.124V` (b) `7.1` (c) calomel electrode

Explanation

At anode: `H_(2) rarr 2H^(+) +2e^(-)`
`E^(@) = - 0.699`
At cathode: `Hg_(2)CI_(2)+2e^(-) rarr 2Hg +2CI^(-)`
`E^(@) = 0.28`
`E_(cell) =- 0.699 +0.28 - (0.0591)/(2) log [H^(+)]^(2) [CI^(-)]^(2)`
`E_(cell) =- 0.419 +0.0591 pH`
(a) `E_(cell) =- 0.419 =0.0591 xx 5`
`=- 0.1235V`
(b) `O =- 0.419 +0.0591 pH`
`pH = 7.1`
(c) `E_(cell) =- 0.419 +0.0591 xx 7.5 = 0.2425` is `E_(Cell) gt 0` cell is spontaneous & calomal electrode with work as cathode i.e. positive electorde.
`(E_(cell) =- 0.419 +0.0591 xx 7.5 = 0.2425 E_(cell) gt 0)`