Calculate the potential of the following cell `:`
`Pt(s)|{:(Ce^(3+),(2M)) ,(Ce^(4+),(1M)):}||{:(Cr^(3+)(2M),) ,(Cr_(2)O_(7)^(2-)(1M),H^(o+)(1M)):}|Pt(s)`
Given `:E^(c-)._(Ce^(3+)|Ce^(4+))=-1.7,E^(c-)._(Cr_(2)O_(7)^(2-)|Cr^(3+))=1.3V`
`(`Take`0.059~~0.06)`

Explanation

`E^(c-)._(Ce^(3+)|Ce^(4+))=-1.7V`
`(` Standard oxidation potential value `)`.
So, `E^(c-)._(Ce^(4+)|Ce^(3+))=1.7V`
`(` Standard reduction potential value `).`
and `E^(c-)._((Cr_(2)O_(7)^(2-)|Cr^(3+)))=1.3V`
`(` Standard reduction potential value `)`.
Anode reduction `:`
`6Ce^(3+)(2M) rarr 6Ce^(4+)(1M)+cancel(6e^(-))`
Cathode reaction `:`
`14H^(o+)(1M)+Cr_(2)O_(7)^(2-)(2M)+cancel(6e^(-))rarr2Cr^(3+)(1M)+7H_(2)O`
Cell reaction `:`
`ulbar(6Ce^(3+)(2M)+14H^(o+)(1M)+Cr_(2)O_(7)^(2-)(2M)rarr 6Ce^(4+)(1M)+7H_(2)O)`
`E^(c-)._(cell)=(E^(c-)._(reduction))_(c)-(E^(c-)._(reduction ))_(a)`
`=1.3V-1.7V=-0.4V`
`E_(cell)=E^(c-)._(cell)-(0.06)/(6)log .([Ce^(4+)]^(6)[Cr^(3+)]^(2))/([Ce^(3+)]^(6)[H^(o+)]^(14)[Cr_(2)O_(7)^(2-)])`
`=-0.4V-0.01log.((1)^(6)(1)^(2))/((2)^(6)(1)^(14)(2))`
`=-0.4V-0.01(log2^(-7))`
`=-0.4V-0.01(-7 log 2)`
`=(-0.4+0.01xx7xx0.3)V`
`=(-0.4+0.021)V=-0.379~~-0.38V`