For the following electrochemical cells, liquid junction potentials affect the cell potential ($E_{cell}$).
I. $\mathrm{Electrode\text{-}1 \mid 0.1\ M\ KCl \parallel 0.1\ M\ NaCl \mid Electrode\text{-}2}$
II. $\mathrm{Electrode\text{-}1 \mid 0.1\ M\ HCl \parallel 0.01\ M\ HCl \mid Electrode\text{-}2}$
III. $\mathrm{Electrode\text{-}1 \mid 0.01\ M\ KCl \parallel 0.01\ M\ HCl \mid Electrode\text{-}2}$
The correct order of their $E_{cell}$ at $25^{\circ}\mathrm{C}$ is

Question

Alkyl halide and aryl halide react in the presence of sodium metal and dry ether. The name of the reaction is?

Answer 1

A quicker route to the same ranking looks at WHICH ion is out of balance at each junction and how badly, without writing the full Henderson formula from scratch each time.

All three cells share the same two electrodes, so differences in $E_{cell}$ come only from the liquid junction potential $E_j$ set up where the two solutions meet. A junction potential arises whenever a cation and an anion diffuse across the boundary at different rates, leaving a small charge imbalance.

Junction II (0.1 M HCl against 0.01 M HCl): the electrolyte is the same on both sides but ten times more concentrated on the Electrode-1 side, so $\mathrm{H^+}$ and $\mathrm{Cl^-}$ both diffuse from the concentrated to the dilute side. $\mathrm{H^+}$ (limiting conductance about 350 $\mathrm{S\,cm^2\,mol^{-1}}$) is far faster than $\mathrm{Cl^-}$ (about 76), so $\mathrm{H^+}$ races ahead, the dilute side becomes momentarily positive, and this gives the largest junction potential of the three, about $+38$ mV.
Junction I (0.1 M KCl against 0.1 M NaCl, same concentration): here $\mathrm{Cl^-}$ is common to both sides so it does not contribute a net gradient; the only mismatch is between $\mathrm{K^+}$ (about 73.5) and $\mathrm{Na^+}$ (about 50.1), which are much closer to each other than $\mathrm{H^+}$ and $\mathrm{Cl^-}$ are. This gives a small positive junction potential, about $+4.4$ mV, which is why KCl and NaCl are considered a comparatively "well matched" pair for this kind of boundary.
Junction III (0.01 M KCl against 0.01 M HCl, same concentration): again $\mathrm{Cl^-}$ is common, so the mismatch is between $\mathrm{K^+}$ (about 73.5) and $\mathrm{H^+}$ (about 350), which is an even bigger mismatch than in junction I. Working through the same Henderson-type expression gives about $-27$ mV, and the sign comes out negative because the very mobile $\mathrm{H^+}$ now sits on the Electrode-2 side, which pushes the potential the opposite way to junction I.

Putting the three signed numbers side by side, $+38.0$ mV (II), $+4.4$ mV (I), and $-27$ mV (III), the order from highest to lowest is II, then I, then III.

Let's summarize:

Same electrolyte, different concentration, with $\mathrm{H^+}$ involved (junction II) gives by far the biggest junction potential.
Same concentration, common anion, different cations (junctions I and III) give smaller potentials, sized by how different the two cation mobilities are, and the sign depends on which side the faster ion sits.

So the correct order is $E_{cell}(\mathrm{II}) > E_{cell}(\mathrm{I}) > E_{cell}(\mathrm{III})$, option (B).

Answer 2

A quicker route to the same ranking looks at WHICH ion is out of balance at each junction and how badly, without writing the full Henderson formula from scratch each time.

All three cells share the same two electrodes, so differences in $E_{cell}$ come only from the liquid junction potential $E_j$ set up where the two solutions meet. A junction potential arises whenever a cation and an anion diffuse across the boundary at different rates, leaving a small charge imbalance.

Junction II (0.1 M HCl against 0.01 M HCl): the electrolyte is the same on both sides but ten times more concentrated on the Electrode-1 side, so $\mathrm{H^+}$ and $\mathrm{Cl^-}$ both diffuse from the concentrated to the dilute side. $\mathrm{H^+}$ (limiting conductance about 350 $\mathrm{S\,cm^2\,mol^{-1}}$) is far faster than $\mathrm{Cl^-}$ (about 76), so $\mathrm{H^+}$ races ahead, the dilute side becomes momentarily positive, and this gives the largest junction potential of the three, about $+38$ mV.
Junction I (0.1 M KCl against 0.1 M NaCl, same concentration): here $\mathrm{Cl^-}$ is common to both sides so it does not contribute a net gradient; the only mismatch is between $\mathrm{K^+}$ (about 73.5) and $\mathrm{Na^+}$ (about 50.1), which are much closer to each other than $\mathrm{H^+}$ and $\mathrm{Cl^-}$ are. This gives a small positive junction potential, about $+4.4$ mV, which is why KCl and NaCl are considered a comparatively "well matched" pair for this kind of boundary.
Junction III (0.01 M KCl against 0.01 M HCl, same concentration): again $\mathrm{Cl^-}$ is common, so the mismatch is between $\mathrm{K^+}$ (about 73.5) and $\mathrm{H^+}$ (about 350), which is an even bigger mismatch than in junction I. Working through the same Henderson-type expression gives about $-27$ mV, and the sign comes out negative because the very mobile $\mathrm{H^+}$ now sits on the Electrode-2 side, which pushes the potential the opposite way to junction I.

Putting the three signed numbers side by side, $+38.0$ mV (II), $+4.4$ mV (I), and $-27$ mV (III), the order from highest to lowest is II, then I, then III.

Let's summarize:

Same electrolyte, different concentration, with $\mathrm{H^+}$ involved (junction II) gives by far the biggest junction potential.
Same concentration, common anion, different cations (junctions I and III) give smaller potentials, sized by how different the two cation mobilities are, and the sign depends on which side the faster ion sits.

So the correct order is $E_{cell}(\mathrm{II}) > E_{cell}(\mathrm{I}) > E_{cell}(\mathrm{III})$, option (B).

Show Hint

The Correct Option is B

Solution and Explanation

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