JEE MAIN - Chemistry (2024 - 8th April Evening Shift - No. 8)

The emf of cell $$\mathrm{Tl}\left|\underset{(0.001 \mathrm{M})}{\mathrm{Tl}^{+}}\right| \underset{(0.01 \mathrm{M})}{\mathrm{Cu}^{2+}} \mid \mathrm{Cu}$$ is $$0.83 \mathrm{~V}$$ at $$298 \mathrm{~K}$$. It could be increased by :
increasing concentration of $$\mathrm{Tl}^{+}$$ ions
increasing concentration of $$\mathrm{Cu}^{2+}$$ ions
increasing concentration of both $$\mathrm{Tl}^{+}$$ and $$\mathrm{Cu}^{2+}$$ ions
decreasing concentration of both $$\mathrm{Tl}^{+}$$ and $$\mathrm{Cu}^{2+}$$ ions

Explanation

To determine how the emf of the cell, $$\mathrm{Tl}\left|\underset{(0.001 \mathrm{M})}{\mathrm{Tl}^{+}}\right| \underset{(0.01 \mathrm{M})}{\mathrm{Cu}^{2+}} \mid \mathrm{Cu}$$, can be increased, we can use the Nernst equation. The Nernst equation for this electrochemical cell is given by:

$$ E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0591}{n} \log \left( \frac{[\mathrm{Tl}^{+}]}{[\mathrm{Cu}^{2+}]} \right) $$

where:

  • $$ E_{\text{cell}} $$ is the emf of the cell.
  • $$ E_{\text{cell}}^\circ $$ is the standard emf of the cell.
  • $$ n $$ is the number of moles of electrons transferred in the cell reaction; here, $$ n = 2 $$.
  • $$ [\mathrm{Tl}^{+}] $$ is the concentration of thallium ions.
  • $$ [\mathrm{Cu}^{2+}] $$ is the concentration of copper ions.

Given that the emf of the cell can be expressed in terms of the concentration of the ions involved, we can see that increasing the concentration of $$[\mathrm{Cu}^{2+}]$$ or decreasing the concentration of $$[\mathrm{Tl}^{+}]$$ will affect the logarithmic term in the Nernst equation:

$$ E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0591}{2} \log \left( \frac{0.001}{0.01} \right) $$

To increase the emf ($$ E_{\text{cell}} $$) of the cell, it is beneficial to have a less negative (or more positive) correction term. This can be achieved by:

  • Increasing the concentration of $$[\mathrm{Cu}^{2+}]$$ ions: This makes the term inside the logarithm smaller (since we are dividing by a larger number), which in turn makes the logarithmic term less negative.

Therefore, the correct answer is:

Option B: increasing concentration of $$\mathrm{Cu}^{2+}$$ ions

Comments (0)

Advertisement