Question

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

• A. I > II > III
• B. II > I > III
• C. III = I > II
• D. II > I = III

Hint:

Rank the signed Henderson liquid-junction potential for each pair; H+ vs Cl- mismatch (junction II) dominates, and check the SIGN, not just the size, for junction III.

Answer 1

The Correct Option is B

Step 1: Identify what actually varies between the three cells.
In each cell, Electrode-1 and Electrode-2 are the same pair of electrodes throughout; only the pair of solutions across the salt junction (\(\parallel\)) changes. So the electrode contribution to \(E_{cell}\) is the same in all three cells, and the ONLY thing that changes \(E_{cell}\) from one cell to another is the liquid junction potential \(E_j\) generated at that salt bridge. Ranking \(E_{cell}\) is therefore the same as ranking the (signed) \(E_j\) of the three junctions.

Step 2: Recall the Henderson equation for a liquid junction.
For ions of charge \(z_i\), mobility \(u_i\) and concentration \(c_i\) on the two sides \(\alpha\) (Electrode-1 side) and \(\beta\) (Electrode-2 side),
\[ E_j = -\frac{RT}{F}\left(\frac{\sum z_i u_i \Delta c_i}{\sum z_i^2 u_i \Delta c_i}\right)\ln\!\left(\frac{\sum z_i^2 u_i c_i(\beta)}{\sum z_i^2 u_i c_i(\alpha)}\right) \]
with \(\Delta c_i = c_i(\beta) - c_i(\alpha)\). We use limiting ionic conductances (proportional to mobility) at \(25^{\circ}\mathrm{C}\): \(\lambda^{\circ}_{H^+}=349.8\), \(\lambda^{\circ}_{K^+}=73.5\), \(\lambda^{\circ}_{Na^+}=50.1\), \(\lambda^{\circ}_{Cl^-}=76.3\) (\(\mathrm{S\ cm^2\ mol^{-1}}\)).

Step 3: Junction I (0.1 M KCl \(\parallel\) 0.1 M NaCl).
The anion \(\mathrm{Cl^-}\) has the same concentration on both sides (no gradient), so only the cations change: \(\mathrm{K^+}\) disappears and \(\mathrm{Na^+}\) appears. Carrying through the Henderson equation for this "common anion, equal total concentration" case reduces it to
\[ E_j(\mathrm{I}) = -\frac{RT}{F}\ln\left(\frac{\lambda_{Na^+}+\lambda_{Cl^-}}{\lambda_{K^+}+\lambda_{Cl^-}}\right) = -0.0257\ln\left(\frac{50.1+76.3}{73.5+76.3}\right) = -0.0257\ln(0.844) \]
\[ E_j(\mathrm{I}) = -0.0257\times(-0.170) \approx +4.4\ \mathrm{mV} \]

Step 4: Junction II (0.1 M HCl \(\parallel\) 0.01 M HCl).
This is the same electrolyte diluted 10-fold, so the standard concentration-cell form applies:
\[ E_j(\mathrm{II}) = \frac{RT}{F}(t_+-t_-)\ln\frac{c_1}{c_2}, \qquad t_+ = \frac{\lambda_{H^+}}{\lambda_{H^+}+\lambda_{Cl^-}}=\frac{349.8}{426.1}=0.821,\ t_-=0.179 \]
\[ E_j(\mathrm{II}) = 0.0257\times(0.821-0.179)\times\ln(10) = 0.0257\times0.642\times2.303 \approx +38.0\ \mathrm{mV} \]
This is large because \(\mathrm{H^+}\) moves far faster than \(\mathrm{Cl^-}\) (proton hopping), so the two ions race across the junction at very different rates, on top of a 10-fold concentration gradient.

Step 5: Junction III (0.01 M KCl \(\parallel\) 0.01 M HCl).
Again \(\mathrm{Cl^-}\) concentration is unchanged across the junction (0.01 M on both sides), so only the cation identity changes, this time from \(\mathrm{K^+}\) to \(\mathrm{H^+}\):
\[ E_j(\mathrm{III}) = -\frac{RT}{F}\ln\left(\frac{\lambda_{H^+}+\lambda_{Cl^-}}{\lambda_{K^+}+\lambda_{Cl^-}}\right) = -0.0257\ln\left(\frac{349.8+76.3}{73.5+76.3}\right) = -0.0257\ln(2.845) \]
\[ E_j(\mathrm{III}) = -0.0257\times1.046 \approx -26.9\ \mathrm{mV} \]
The sign is negative here (opposite to I) because the fast-moving \(\mathrm{H^+}\) is now the ion appearing on the Electrode-2 side rather than disappearing from the Electrode-1 side, which reverses the direction of net charge separation at the junction.

Step 6: Rank the three signed values.
\[ E_j(\mathrm{II})\approx +38.0\ \mathrm{mV} \ >\ E_j(\mathrm{I})\approx +4.4\ \mathrm{mV} \ >\ E_j(\mathrm{III})\approx -26.9\ \mathrm{mV} \]
So \(E_{cell}(\mathrm{II}) > E_{cell}(\mathrm{I}) > E_{cell}(\mathrm{III})\). Options (A), (C) and (D) all mis-rank at least one of these signed values (in particular they treat III as comparable to I, or place I above II), so they are wrong.

Final Answer:
\[ \boxed{\mathrm{II > I > III}} \], option (B).