Question

The emf of the below cell is:
$\text{Pt,}{{\text{H}}_{\text{2}}}\left( {{\text{P}}_{\text{1}}} \right)\left| \text{HCl} \right|{{\text{H}}_{\text{2}}}\left( {{\text{P}}_{\text{2}}} \right)\text{,Pt}$

A. $\dfrac{\text{RT}}{\text{F}}\text{log}\dfrac{{{\text{P}}_{\text{1}}}}{{{\text{P}}_{\text{2}}}}$
B. $\dfrac{\text{RT}}{\text{2F}}\text{log}\dfrac{{{\text{P}}_{\text{1}}}}{{{\text{P}}_{\text{2}}}}$
C. $\dfrac{\text{2RT}}{\text{F}}\text{log}\dfrac{{{\text{P}}_{\text{1}}}}{{{\text{P}}_{\text{2}}}}$
D. none of the above

Answer 1

Hint: In the given electrode, we must remember to write the two half reactions whose addition will give us the overall cell reaction, and thus we will be able to understand the redox reaction. And to express the electrode reaction, we will take the help of Nernst equation.

Complete step by step solution:
Now, let's try to understand here few terms related to electrochemistry:
${{\text{E}}_{\text{cell}}}$ - known as the emf of the cell, it is defined as the maximum potential difference between two electrodes of a cell.
${{\text{E}}^{\circ }}_{\text{cell}}$- this is the standard electrode potential of the cell at 298 K.
Now, let's try to solve the given numerical above:
- Here the ${{\text{E}}_{\text{cell}}}$ will be the measure of potential difference between the two half cells of the electrode, which as hydrogen in both anode and cathode, and platinum rods.
- The half reaction at cathode and anode are:
In the left hand side cell(L.H.S) : it undergoes reduction and thus behaves as anode.
In the right hand side (R.H.S): it undergoes oxidation, thus behaves as cathode.
Since we have platinum rods, which serves as an inert electrode, it will only provide the surface for the electrons and it's conduction without taking part in the chemical reaction .
-L.H.S half cell:${{\text{H}}_{\text{2}}}\to \text{2}{{\text{H}}^{\text{+}}}\text{+2}{{\text{e}}^{\text{-}}}$ $\left( {{\text{P}}_{\text{1}}} \right)$
R.H.S half cell:$\text{2}{{\text{H}}^{\text{+}}}\text{+2}{{\text{e}}^{\text{-}}}\to {{\text{H}}_{\text{2}}}$ $\left( {{\text{P}}_{\text{2}}} \right)$
Net reaction:$\left( {{\text{P}}_{\text{1}}} \right){{\text{H}}_{\text{2}}}\to {{\text{H}}_{\text{2}}}\text{(}{{\text{P}}_{\text{2}}}\text{)}$
Now, as we talking about the hydrogen atom, whose standard electrode potential is 0(${{\text{E}}^{\circ }}_{\text{cell}}$ )
- Therefore the emf of a cell will be given by the formula: ${{\text{E}}_{\text{cell}}}$ = ${{\text{E}}^{\circ }}_{\text{cell}}$ $\text{ }\\!\\!~\\!\\!\text{ -RTlogQ}$
Or the other formula can also be written as: ${{\text{E}}_{\text{cell}}}\text{=}{{\text{E}}^{\text{o}}}_{\text{cell}}\text{-}\dfrac{\text{RT}}{\text{nF}}\text{log}\dfrac{{{\text{P}}_{\text{1}}}}{{{\text{P}}_{\text{2}}}}$
Where ${{\text{P}}_{\text{1}}}$ is the concentration of cathode and ${{\text{P}}_{2}}$ is of anode respectively.
Here as two electrons are used up in the process therefore n=2
This is also known as Nernst equation.
Where R is gas constant,
T is temperature,
F is the faraday constant -96487 $\text{Cmo}{{\text{l}}^{\text{-1}}}$
And n is the net number of electrons participating in the redox reaction.
- Now as we know ${{\text{E}}^{\circ }}_{\text{cell}}$ = 0
-Therefore after putting the values in the equation we get:${{\text{E}}_{\text{cell}}}=\dfrac{\text{RT}}{\text{2F}}\text{log}\dfrac{{{\text{P}}_{\text{1}}}}{{{\text{P}}_{\text{2}}}}$
So the correct option here is: B

Note: It is seen that that ${{\text{E}}^{\circ }}_{\text{cell}}$ depends upon the concentration of both the ions present in the cathode and anode. While it increases with the increase in concentration of ions present in cathode and decreases with the concentration of ions present in anode.