Question

In the following question, more than one answer may be correct.
Select the correct answers and make it according to the code:
In the cell ${ Zn(s)|Zn }^{ +2 }{ \parallel H }^{ + }{ |H }_{ 2 }{ Pt }$ the addition of ${ H }_{ 2 }{ SO }_{ 4 }$ to the cathode compartment will:
1. Decrease E
2. Increase E
3. Shift equilibrium to left
4. Shift equilibrium to the right

A. 1, 2 and 3 are correct.
B. 1 and 2 are correct
C. 2 and 4 are correct
D. 1 and 3 are correct

Answer 1

Le Chatelier’s principle states that “if a change is introduced in a system at equilibrium in terms of concentration, temperature or pressure, then the system will tend to shift its equilibrium position to counterbalance or neutralize the effect of the change.”

Complete answer:
${ Zn(s)|Zn }^{ +2 }{ \parallel H }^{ + }{ |H }_{ 2 }{ Pt }$
The following reactions will take place:
At anode: ${ Zn\rightarrow Zn }^{ +2 }{ +2e }^{ - }$
At cathode: ${ 2H }^{ + }{ +2e }^{ - }{ \rightarrow H }_{ 2 }$
So, the net reaction will be: ${ Zn+2H }^{ + }{ \rightarrow { Zn }^{ +2 }+H }_{ 2 }$
According to the Nernst equation:
${ E }_{ cell }{ =E }_{ cell }^{ \circ }{ -RT/nFlogQ }$
where ${ E }_{ cell }$ = cell potential at non-standard state conditions
${ E }_{ cell }^{ \circ }$ = electrode cell potential at standard conditions
R = Universal gas constant
T = temperature in kelvin
n = number of moles of electron transferred in a reaction
F = Faraday’s constant
Q = reaction quotient
Now, ${ E }_{ cell }{ =E }_{ cell }^{ \circ }{ =-RT/nFlog }{ [Zn }^{ +2 }{ ]/(H }^{ + }{ ) }^{ 2 }{ ] }$ ........(1)
${ E }_{ cell }{ =E }_{ cell }^{ \circ }{ =-0.0591/2 log }{ [Zn }^{ +2 }{ ]/(H }^{ + }{ ) }^{ 2 }{ ] }$...........(2)
${ E }_{ cell }{ =E }_{ cell }^{ \circ }{ =+0.0591/2log }[{ ]/(H }^{ + }{ ) }^{ 2 }/{ Zn }^{ +2 }{ ] }$...........(3)
According to Le Chatelier’s principle, in the net reaction, when the concentration of ${ H }^{ + }$ ion increases, equilibrium shifts in the right direction. According to the Nernst equation, when the ${ H }^{ + }$ ion concentration increases, the value ${ E }_{ cell }$ will increase.
In the cell${ Zn(s)|Zn }^{ +2 }{ \parallel H }^{ + }{ |H }_{ 2 }{ Pt }$ the addition of ${ H }_{ 2 }{ SO }_{ 4 }$ to the cathode compartment (towards the reactant side) then the reaction shifts in forwarding direction,i.e, towards the right and ${ E }_{ cell }$ increases.
Hence, option C is correct.

Note: The possibility to make a mistake is that in the Nernst equation we can clearly see that ${ E }_{ cell }$ is inversely proportional to the ${ H }^{ + }$ ion concentration, but when the negative sign changes into positive then the will be directly proportional to the ${ H }^{ + }$ ion concentration, to ${ H }^{ + }$ ion concentration increases, E increases.