electrochemical cell..pp

The solubility of AgCl is very low. The [Ag+(aq)] in the right half cell is much smaller than 0.01 M. Therefore, the reaction that occurs in the right half cell is:
Ag(s) → Ag+(aq) + e- ...E = -ER
ER and EL are defined as reduction potential. Since oxidation occurs in the right half cell, thus the electrode potential of the right half cell is -ER (ER is negative).

The Ag+(aq) in the right half cell is formed from the dissociation of AgCl(s):
AgCl(s) ≒ Ag+(aq) + Cl-(aq) ...... (*)

When KCl(aq) is added, [Cl-(aq)] increase and this shifts the equilibrium position of reaction (*) to the left. Therefore, [Ag+(aq)] decreases. This shifts the equilibrium position of the first reaction to the right. Ag(s) has a greater tendency to change to Ag+(aq), and thus E increases and ER decreases (more negative).
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