The `EMF` of a galvanic cell `Pt|H_(2)(1 atm)|HCl(1M)|Cl_(2)(g)|Pt` is `1.29V`. Calculate the partial pressure of` Cl_(2)(g)`. `E^(c-)._(Cl_(2)|Cl^(c-))=1.36V`.

Explanation

Cell reaction `:`
`(1)/(2)H_(2)(g)rarr 2H^(o+)+e^(-)" "E^(c-)._(o x i d)=0V`
`2Cl_(2)(g)+e^(-) rarr Cl^(c-)(aq)" "E^(c-)._(red)=1.36V`
`ulbar((1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)rarrH^(o+)(aq)+Cl^(c-)(aq)" "E^(c-)._(cell)=1.36V)`
`E_(cell)=E^(c-)._(cell)-(0.059)/(1) log.([H^(o+)][Cl^(c-)])/([P_(H_(2))]^(1//2)[P_(Cl_(2))]^(1//2))`
`1.29=1.36-(0.059)/(1)log ((1xx1)/(1xx[p_(Cl_(2))]^(1//2)))`
Solve, `p_(Cl_(2))=4.24xx10^(-3)atm`