Solid NaCl is added to the solution of a Ag+/Ag half cell until AgCl precipitates. The electrode potential

decreases, because [Ag+] in solution falls

increases, because Cl- is added

remains unchanged, because NaCl is not involved in the half reaction

becomes zero, because AgCl is insoluble

NEET levelSingle correct optionDifficulty 1 of 3ec100

Solid NaCl is added to the solution of a $Ag X^{+} / Ag$ half cell until AgCl precipitates. The electrode potential

E = E^{\circ} (Ag X^{+} / Ag) + 0.0591 lo g [Ag X^{+}]

. Precipitation of AgCl removes

Ag X^{+}

from solution, so

lo g [Ag X^{+}]

becomes more negative and

E

decreases. The electrode now behaves as a

Ag ∣ AgCl ∣ Cl X^{-}

electrode with

E^{\circ} = 0.22 V

, well below 0.80 V, but it is not zero.