For the cell $Pt ∣ H X_{2} (1 bar) ∣ HCl (m) ∣ AgCl (s) ∣ Ag$ , treating activities as molalities, a ten-fold increase in $m$ changes the cell EMF by

Answer: B

Cell reaction:

\frac{1}{2} H X_{2} + AgCl Ag + H X^{+} + Cl X^{-}

n = 1

Q = [H X^{+}] [Cl X^{-}] = m^{2}

. Nernst:

E = E^{\circ} - 0.0591 lo g m^{2} = E^{\circ} - 0.1182 lo g m

. Increasing

m

ten-fold lowers

E

0.118 V

. The two electrodes share the electrolyte but the product ions still appear in

Q