InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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» `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. |
|