Solveeit Logo

Question

Question: The oxidation potential values of \(A,B,C\) and \(D\) are \(-0.03V,\ +0.108V,\ -0.07V\) and \(+0.1V\...

The oxidation potential values of A,B,CA,B,C and DD are 0.03V, +0.108V, 0.07V-0.03V,\ +0.108V,\ -0.07V and +0.1V+0.1V respectively. The non-spontaneous cell reaction takes place between them:
a.A&Ba.\,\,A\And B
b.B&Db.\,\,B\And D
c.D&Ac.\,\,D\And A
d.B&Cd.\,\,B\And C

Explanation

Solution

There are two types of reactions namely spontaneous and non-spontaneous. A reaction is spontaneous when the oxidation potential value of one is higher than the other, in that case one with higher value will act as anode and other with lower value as cathode.

Complete step by step answer:
(i) The electrode with higher oxidation potential will act as anode and electrode with lower oxidation potential will act as cathode.
(ii) Cell reaction is spontaneous when Ecell{{E}_{cell}} is positive and it is given as
Ecell=ECEA{{E}_{cell}}={{E}_{C}}-{{E}_{A}} where EC{{E}_{C}} is the potential of cathode and EA{{E}_{A}} is the potential of anode
So, if we get ECell>0{{E}_{Cell}}>0 that is positive then reaction is spontaneous
Otherwise if ECell<0{{E}_{Cell}}<0 that is negative then reaction is non- spontaneous
So, we will take each option one by one and compare the potential values and find the spontaneity of the reaction
Now, we are given,
Oxidation potential of A= 0.03VA=\ -0.03V
Oxidation potential of B= +0.108VB=\ +0.108V
Oxidation potential of C=0.07VC=\,-0.07V
Oxidation potential of D= +0.1VD=\ +0.1V
(a)\left( a \right) AA and BB, oxidation potential of A= 0.03A=\ -0.03
Oxidation potential of B= +0.108B=\ +0.108
\because Oxidation potential of BB is more than A.A.
\because Oxidation of AA is not possible, and the cell is not possible in this case.
\because This cell having AA as anode and BB as cathode is not possible.
(b)\left( b \right) B&D,B\And D, oxidation potential B=+0.108VB=+0.108V and D=+0.1VD=+0.1V
\because oxidation potential of BB is more than DD.
\therefore oxidation of BB is possible and cell reaction is feasible.
(c)\left( c \right) D&A,D\And A, oxidation potential of D=+0.1V&A=0.03VD=+0.1V\And A=-0.03V
\because oxidation potential of DD is more than AA
\therefore cell will work with DD as anode and AA as cathode.
(d)\left( d \right) B&C,B\And C, oxidation potential of B=+0.108VB=+0.108V and C=0.07VC=-0.07V
\because oxidation potential of BB is more than C.C.
\therefore cell will work with BB as anode and CC as cathode.
\therefore The non-spontaneous cell reaction takes place between AAand BB

So, the correct answer is Option A.

Additional Information:
The second law of thermodynamics states that for any spontaneous process, the overall gas must be greater than or equal to zero, yet, spontaneous chemical reactions can result in a negative change in entropy.
The ΔS\Delta S of the surroundings increases enough because of the exothermicity of the reaction so that it overcompensates for the negative ΔS\Delta S of the system. Since the overall
ΔS=ΔSsurroundings+ΔSsystem\Delta S=\Delta {{S}_{surroundings}}+\Delta {{S}_{system}}
The overall change in entropy is still positive.

Note: The reaction will be spontaneous in the reverse direction and non-spontaneous redox reaction cannot be used to produce electricity.
If E(redox reaction){{E}^{\circ }}_{(redox\ reaction)} is negative
(E(redox reaction)<0),\left( {{E}^{\circ }}_{(redox\ reaction)}<0 \right), the reaction will not proceed in the forward direction (non-spontaneous).