Question

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

Answer 1

Hint: 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 ${{E}_{cell}}$ is positive and it is given as
${{E}_{cell}}={{E}_{C}}-{{E}_{A}}$ where ${{E}_{C}}$ is the potential of cathode and ${{E}_{A}}$ is the potential of anode
So, if we get ${{E}_{Cell}}>0$ that is positive then reaction is spontaneous
Otherwise if ${{E}_{Cell}}<0$ 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 $A=\ -0.03V$
Oxidation potential of $B=\ +0.108V$
Oxidation potential of $C=\,-0.07V$
Oxidation potential of $D=\ +0.1V$
\[\left( a \right)\] $A$ and $B$, oxidation potential of $A=\ -0.03$
Oxidation potential of $B=\ +0.108$
$\because $ Oxidation potential of $B$ is more than $A.$
$\because $ Oxidation of $A$ is not possible, and the cell is not possible in this case.
$\because $ This cell having $A$ as anode and $B$ as cathode is not possible.
\[\left( b \right)\] $B\And D,$ oxidation potential $B=+0.108V$ and $D=+0.1V$
$\because $ oxidation potential of $B$ is more than $D$.
$\therefore $ oxidation of $B$ is possible and cell reaction is feasible.
\[\left( c \right)\] $D\And A,$ oxidation potential of $D=+0.1V\And A=-0.03V$
$\because $ oxidation potential of $D$ is more than $A$
$\therefore $ cell will work with $D$ as anode and $A$ as cathode.
\[\left( d \right)\] $B\And C,$ oxidation potential of $B=+0.108V$ and $C=-0.07V$
$\because $ oxidation potential of $B$ is more than $C.$
$\therefore $ cell will work with $B$ as anode and $C$ as cathode.
$\therefore $ The non-spontaneous cell reaction takes place between \[A\]and \[B\]

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 $\Delta S$ of the surroundings increases enough because of the exothermicity of the reaction so that it overcompensates for the negative $\Delta S$ of the system. Since the overall
$\Delta S=\Delta {{S}_{surroundings}}+\Delta {{S}_{system}}$
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 ${{E}^{\circ }}_{(redox\ reaction)}$ is negative
$\left( {{E}^{\circ }}_{(redox\ reaction)}<0 \right),$ the reaction will not proceed in the forward direction (non-spontaneous).