Discussion
For the following electrochemical cells, liquid junction potentials affect the cell potential (Ecell).
I. Electrode-1 | 0.1 M KCl || 0.1 M NaCl | Electrode-2
II. Electrode-1 | 0.1 M HCl || 0.01 M HCl | Electrode-2
III. Electrode-1 | 0.01 M KCl || 0.01 M HCl | Electrode-2
The correct order of their Ecell at 25 °C is
Answer: B
The order is \(\mathrm{II} \gt \mathrm{I} \gt \mathrm{III}\).
The question attributes the differences between the cells to their liquid junctions, so compare the junction potential each adds, taken as the potential of the right-hand solution relative to the left. A junction charges the side towards which the faster ion runs.
- II (0.1 M HCl | 0.01 M HCl): H+ (transport number about 0.82) diffuses into the dilute side much faster than Cl-, making the right side positive by roughly \(0.64 \times 59\) mV, about +38 mV.
- I (0.1 M KCl | 0.1 M NaCl): common anion, and K+ is only slightly more mobile than Na+, so the right side is a few millivolts positive.
- III (0.01 M KCl | 0.01 M HCl): H+ runs leftwards into the KCl far faster than K+ runs right, making the left side positive, a negative contribution of roughly -27 mV.
So Ecell falls in the order II, I, III. The options treating I and III as equal ignore the large biionic junction in III, and \(\mathrm{I} \gt \mathrm{II} \gt \mathrm{III}\) misses that the tenfold HCl gradient in II gives the largest junction of all.
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