Question

Calculate EMF of the cell:
$Pt(\left. {{H_2})} \right|{H^ + }\left. {(0.1M)} \right\|{H^ + }\left. {(0.2M)} \right|Pt({H_2})$

Answer 1

We have to use the Nernst equation in this kind of question. The Nernst equation is a very simple equation that helps in determining the EMF of any given cell. All we need are the standard EMF of the cell and the concentrations and temperature of the system. In this case, we are using the same electrolytes and therefore the standard EMF will be 0.

Complete answer:
We know that this is an electrolyte concentration cell with ${H^ + }$ as the electrolyte. According to Nernst equation:
$\Rightarrow {E_{cell}} = E_{cell}^0 - \dfrac{{RT}}{{nF}}\log Q$
Here ${E_{cell}}$ is the EMF of the cell
$E_{cell}^0$ is the standard EMF of the cell
R is the universal gas constant
T is the temperature
n is the number of moles of electron used in the reaction
F is the Faraday's constant
Q is the reaction quotient
We can say that since both the electrolytes in this reaction is ${H^ + }$ , the standard EMF of the cell is 0.
Therefore the Nernst equation becomes:
$\Rightarrow {E_{cell}} = \dfrac{{2.303RT}}{{nF}}\log Q$
Q can be written in terms of the concentration of ${H^ + }$ ions and thu we can rewrite the above equation as:
$\Rightarrow {E_{cell}} = \dfrac{{2.303RT}}{{nF}}\log \dfrac{{{C_2}}}{{{C_1}}}$
Where ${C_2}$ and ${C_1}$ are the concentrations represented in the following equation:
$\Rightarrow {H^ + }({C_2}) \rightleftharpoons {H^ + }({C_1})$
By substituting the values given we get:
$\Rightarrow {E_{cell}} = \dfrac{{2.303 \times 8.314 \times 298}}{{1 \times 96500}}\log \dfrac{{0.2}}{{0.1}}$
We can see that the simplification of the log term will give us log 2 which is 0.3010. We can substitute that in the equation and get:
$\Rightarrow {E_{cell}} = \dfrac{{2.303 \times 8.314 \times 298}}{{1 \times 96500}} \times 0.3010$
By simplifying the terms left to 0.3010, we get:
$\Rightarrow {E_{cell}} = 0.0593 \times 0.3010$
Simplifying this we get:
$\Rightarrow {E_{cell}} = 0.0177{\text{ }}V$
Thus the EMF of the cell is 0.0177 V.

Note:
This question is solved by considering that there is no transference in the concentration cell. Here the half-cells are in different vessels and therefore don’t have any liquid junction potential. Salt bridge will nullify the Liquid junction potential here. In cells with transference, we have to also consider the transport number of the ions.