Question

To find the standard potential of $$M^{3+}/M$$ electrode, the following cell is constituted:$Pt/M/M^{+}(0.001mol{L}^{-1})/A{g}^{+}(0.01mol{L}^{-1})/Ag$ . The emf of the cell is found to be $0.421\text{volt}$ at $298K$ . The standard potential of half reaction $M^{+3}+3e^{-}\to M$ at $298K$ will be:
(Given:${E^e}_{A{g}^{+}/Ag}\text{ at 298K = 0}\text{.80Volt}$)
A.$\text{0}\text{.32 Volt}$
B.$\text{0}\text{.66 Volt}$
C.$\text{1}\text{.28 Volt}$
D.$\text{0}\text{.38 Volt}$

Answer 1

Hint: The device used to generate electricity from chemical reactions or vice versa is known as an electrochemical cell. The standard emf of a cell is the difference of the standard potentials at the cathode and the anode.

Complete step by step answer:
An electrochemical cell is a device used to generate electricity from chemical reactions or use electricity to proceed the chemical reactions.
As we all know that electricity and the potential difference have relation in them similarly here we will see the term cell potential, basically it is the measure of potential difference between the two half cells of an electrochemical cell. It is also represented as ${\text{E}}_{cell}$ . This potential difference is developed when there is flow of electrons from one half cell to another.
When this potential difference is measured under standard conditions, it is termed as standard potential. The standard conditions are :the concentration should be $\text{1M}$ , pressure should be $1\text{atm}$ and the temperature should be ${25}^{\circ }C$ .
In this problem we have to calculate the standard potential of half reaction. The formula to calculate standard potential is:
$E_{cell}=E{}_{cell}^{\circ }-{\displaystyle \frac{0.6}{n}}\log {\displaystyle \frac{\left[M^{+3}\right]}{{\left[A{g}^{+}\right]}^3}}$
Where,
$E_{cell}$is the emf of a cell.
$E{}_{cell}^{\circ }$ is the standard emf of a cell.
$n$ is the number of electrons involved in reaction.

And we have given that $E_{cell}=0.421\text{volt}$,
$\left[M^{+3}\right]=0.001$ and $\left[A{g}^{+}\right]=0.01$. The number of electrons involved in reaction are three so $n=3$ .
So by putting all the values in formula we get
$0.421=E{}_{cell}^{\circ }-{\displaystyle \frac{0.6}{3}}\log {\displaystyle \frac{\left[0.001\right]}{{\left[0.01\right]}^3}}$
By solving we get.
$E{}_{cell}^{\circ }=0.48\text{V}$
Now,
We all know that the standard emf of a cell is the difference of the standard potentials at the cathode and the anode.
Therefore, $E{}_{cell}^{\circ }=E{}_{Ag}^{\circ }-E{}_M^{\circ }$
We have obtained the value of $E{}_{cell}^{\circ }$ which is $0.48\text{volt}$.
And we have given that $E{}_{Ag}^{\circ }=0.80\text{V}$
So by putting all the values we get,
$0.48=0.80-E{}_M^{\circ }$
$\Rightarrow E{}_M^{\circ }=0.80-0.48=0.32\text{V}$
Hence the standard potential of the half reaction $M^{+3}+3e^{-}\to M$ is $\text{0}\text{.32V}$.
So, option (A) is correct.

Note:
The conductivity or conductance of a cell is measured experimentally from the electrical resistance produced by the electrolyte. An alternating voltage is applied across the two electrodes present in electrolyte and ions are allowed to move toward the electrodes and the resistance is noted.