The overall reaction for the passage of 1.0 faraday of charge in the following cell $\mathrm{Ag}(\mathrm{s})-\mathrm{AgCl}(\mathrm{s})\left|\mathrm{KCl}\left(\mathrm{a}_{1}\right)\right| \mathrm{KCl}\left(\mathrm{a}_{2}\right) \mid \mathrm{AgCl}(\mathrm{s})-\mathrm{Ag}(\mathrm{s})$ is given by (t denotes the transport numbers)
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
With Ag/AgCl electrodes reversible to Cl⁻, passage of 1 F moves t₊ mol of KCl across the junction from the more concentrated to the more dilute solution: t₊ KCl(a₂) → t₊ KCl(a₁) (the cell is spontaneous for a₂ > a₁).