For the cell `:`
`Zn (s) |Zn^(2+)(a M)||Ag^(o+)(bM)|Ag(s)`.
`a.` Write Nernst equation to show how `E_(cell)` vary with concentration of `Zn^(2+)` and `Ag^(o+)` ions. Given `E^(c-)._((Zn^(2+)|Zn))=0.76V,E^(-)._((Ag^(o+)|Ag))=0.80V. `
`b.` Find `E_(cell)` for `[Zn^(2+)]=0.01M` and `[Ag^(c-)]=0.02M`.
`c.` For what values of `Q` will the cell `EMF` be
`i. 0.0V" "ii. 0.97 V`

Explanation

`a.`
Anode reaction `: " " Zn(s) rarr Zn^(2+)(aM)+cancel(2e^(-))`
Cathode reaction `: `
`2Ag^(o+)(bM)+cancel(2e^(-)) rarr 2Ag(s)`
`ulbar(Cell reaction:Zn(s)+2Ag^(o+)(bM) rarr Zn^(2+)(aM)+2Ag(s))`
`E^(c-)._(cell)=(E^(c-)._(reduction))_(c)=(E^(c-)._(reduction))_(a)`
`=0.80-(-0.76)=1.56V`
Using Nernst equation,
`E_(cell)=E^(c-)._(cell)-(0.059)/(n_(cell))log.([Zn^(2+)])/([Ag^(o+)])`
`(` Activity of `Zn(s) ` and `Ag(s)=1)`
`=E^(c-)._(cell)-(0.059)/(2)log[Zn^(2+)]+(0.059)/(2)log[Ag^(o+)]^(2)`
Therefore, `E_(cell)` will decrease when `[Zn^(2+)]` increases and `[Ag^(o+)]` decreases.
`b.` iF `[Zn^(2+)]=0.01M` and `[Ag^(o+)]=0.02M`
`E_(cell)=1.56V-(0.059)/(2)log.(0.01)/((0.02)^(2))`
`=1.56V-(0.059)/(2)[log 25]`
`=1.56V-(0.059)/(2)[2 log 5]`
`=1.56 V-(0.059)/(2)xx2xx0.7" "(1 lo ~~ 0.7)`
`=1.56V-0.059xx0.7`
`=1.56-0.0413=1.5187V`
`c.` `i.` `E_(cell=0.0V,E^(c-)._(cell)=1.56V`
`E_(cell)=E^(c-)._(cell)-(0.059)/(2)log Q_(cell)`
`0.0=1.56-(0.059)/(2)log Q_(cell)`
`log Q_(cell)=(2xx1.56)/(0.059)=52.88`
`Q_(cell)=Anilog(52.88=7.586xx10^(52)`
`ii. E_(cell)=0.97V,E^(c-)._(cell)=1.56V`
`E_(cell)=E^(c-)._(cell)-(0.059)/(2)log Q_(cell)`
`0.97=1.56-(0.059)/(2) log Q_(cell)`
`logQ_(cell)=(2xx0.59)/(0.059)=20`
`Q_(cell)=Anilog(20)=10^(20)`