Question

For the cell : `Zn|{:(Zn_(aq.)^(2+)),(1M):}||{:(Cu_(aq.)^(2+)),(2M):}|Cu`
Calculate the values for ,
(a) cell reaction,
(b) `E_(cell)^(@)`
(c) `E_(cell)`
(d) the minimum concentration of `Cu^(2+)` at which cell reaction is spontaneous if `Zn^(2+)` is `1M`,
(e) does the displacement of `Cu^(2+)` goes almost to completion.
Given : `E_(RP_(Cu^(2)//Cu))^(@) = +0.35V`
`E_(RP_(Zn^(2)//Zn))^(@) = -0.76V`

Answer 1

`E_(OP)^(@)` for `Cu//Cu^(2+) = -0.35 V`
`E_(OP)^(@)` for `Zn//Zn^(2+) = +0.76 V`
More is `E_(OP)^(@)`, more is tendency to show oxidation and thus `Zn` will oxidize and `Cu^(2+)` will reduce.
Anode : `Zn rarr Zn^(2+)+2e`
Cathode : `Cu^(2+)+2e rarr Cu`
Cell reaction : `Zn+Cu^(2+) rarr Zn^(2+)+Cu`
(b) Also, `E_(cell)^(@) = E_(OP_(Zn//Zn^(2+)))^(@)+E_(RP_(Cu^(2+)//Cu))^(@)`
`= 0.76+0.35 = 1.11 V`
(c) Also, `E_(cell) = E_(Zn//Zn^(2+))^(@).^(+)E_(Cu^(2+)//Cu)`
`+E_(RP_(Cu//Cu^(2+)))^(@)+(0.059)/(2)log[Cu^(2+)]`
`E_(cell) = E_(OP_(Zn//Zn^(2+)))^(@)+E_(RP_(Cu//Cu^(2+)))^(@)+(0.059)/(2)log.([Cu^(2+)])/([Zn^(2+)])`
`= 0.76 + 0.35 + (0.059)/(2)log.(2)/(1)`
`E_(cell) = 1.109 V`
(d) Also,`E_(cell) = 1.1 + (0.059)/(2)log.([Cu^(2)])/([Zn^(2+)])`
Thus, if `E_(Cell)` is `+ve` for `[Zn^(2+)] = 1`,
so cell reaction is spontaneous.
`(0.059)/(2)log.([Cu^(2)])/([Zn^(2+)])gt -1.1`
or `log.([Cu^(2)])/([Zn^(2+)])gt -(1.1 xx 2)/(0.059) = -37.29`
or `log .([Cu^(2+)])/(1)gt -37.29`
or `[Cu^(2+)] gt 5.13 xx 10^(-38)`
(e) Yes, the displacement almost goes to completion.