Calculate the emf of the cells formed by the various combinations of the following standard half cells. Here [M^(n+)]=1 mol `L^(-1)`, since we are con

Question

Calculate the emf of the cells formed by the various combinations of the following standard half cells. Here [M^(n+)]=1 mol `L^(-1)`, since we are con

Calculate the emf of the cells formed by the various combinations of the following standard half cells. Here [M^(n+)]=1 mol `L^(-1)`, since we are considering standard cells.
(i) `Zn^(2+)(aq)//Zn(s)`
(ii) `Cr^(3+)(aq)//Cr(s)`
(iii) `Cu^(2+)(aq)//Cu(s)`
(IV) `Ni^(2+)(aq)//Ni(s)`
(v) `Co^(2+)(aq)//Co(s)`
(vi) `Ag^(+)(aq)//Ag(s)` also calculate the standard potentials of such cells.

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

The `E^(@)` values (reduction potentials) for the given half -cells obtained from Table 8.3 are given below : `{:("Half-cell " : Zn^(2+)//Zn, Cr^(3+)//Cr, Cu^(2+)//Cu,Ni^(2+)//Ni, Co^(2+)//Co,Ag^(+)//Ag),(E^(@) "value(v) " : -0.76,-0.74,+0.34,-0.25,-0.28,+0.80):}`
The increasing order of `E^(@)` value is :
`Zn^(2+)//Zn lt Cr^(3+)//Cr lt Co^(2+)//Co lt Ni^(2+)//Ni lt Cu^(2+)//Cu lt Ag^(+)//Ag`.
From the data it is evdent that
(a) A cell with `Zn^(2+)//Zn` as anode can have all the five other electrodes acting as cathode. The emf of the cells in all the cases are :
(i)`E_((Cr^(3+)//Cr))^(@)-E_((Zn^(2+)//Zn))^(@)=(-0.74)-(-0.76)=0.02 V`
(ii) (ii)`E_((Co^(2+)//Cu))^(@)-E_((Zn^(2+)//Zn))^(@)=(-0.28)-(-0.76)=0.48 V`
(iii) `E_((Ni^(2+)//Ni))^(@)-E_((Zn^(2+)//Zn))^(@)=(-0.25)-(-0.76)=0.51 V`
(iv) `E_((Cu^(2+)//Cu))^(@)-E_((Zn^(2+)//Zn))^(@)=(+0.34)-(-0.76) = 1.10 V`
(v) `E_((Ag^(+)//Ag))^(@)-E_((Zn^(2+)//Zn))^(@)=(+0.80)-(-0.76) = 1.56 V`
(b) A cell with `Cr^(3+)//Cr` as anode can have all the remaining four electrodes acting as cathode.
The emf of the cells in all the cases are :
(i)`E_((Co^(2+)//Co))^(@)-E_((Cr^(3+)//Cr))^(@)=(-0.28)-(-0.74)=0.46 V`
(ii)`E_((Ni^(2+)//Ni))^(@)-E_((Cr^(3+)//Cr))^(@)=(-0.25)-(-0.74)=0.49 V`
(iii)`E_((Cu^(2+)//Cu))^(@)-E_((Cr^(3+)//Cr))^(@)=(+0.34)-(-0.74)=1.08 V`
(iv)`E_((Ag^(+)//Ag))^(@)-E_((Cr^(3+)//Cr))^(@)=(+0.80)-(-0.74)=1.54 V`
(c ) A cell with `Co^(2+)//Co` as anode can have all the remaining three electrodes acting as cathode.
The emf of the cells can be calculated as :
(i)`E_((Ni^(2+)//Ni))^(@)-E_((Co^(2+)//Co))^(@)=(-0.25)-(-0.28)=0.03 V`
(ii)`E_((Cu^(2+)//Cu))^(@)-E_((Co^(2+)//Co))^(@)=(+0.34)-(-0.28)=0.62 V`
(iii)`E_((Ag^(+)//Ag))^(@)-E_((Co^(2+)//Co))^(@)=(+0.80)-(-0.28)=1.08 V`
(d) A cell with `Ni^(2+)//Ni` as anode can have the remaining two elctrodes acting as the cathode.
The emf of the cells can be calculated as :
(i)`E_((Cu^(2+)//Cu))^(@)-E_((Ni^(2+)//Ni))^(@)=(+0.34)-(-0.25)=0.59 V`
(ii)`E_((Ag^(+)//Ag))^(@)-E_((Ni^(2+)//Ni))^(@)=(+0.80)-(-0.25)=1.05 V`
(e) A cell with `Cu^(2+)..Cu` as anode can have only `Ag^(+)//Ag` as cathode.
(i)`E_((Ag^(+)//Ag))^(@)-E_((Cu^(2+)//Cu))^(@)=(+0.80)-(+0.34)=0.46 V`