In the cell Pt(s)|H2(g,1\ bar)|HCl(aq)|AgCl(s)|Ag(s)|Pt(s), the cell potential is 0.92 V when a 10-6 molar HCl solution…

Strategy: For the silver-silver chloride electrode, the potential depends on the concentration of $\ce{Cl-}$. In an $\ce{HCl}$ solution, both $\ce{H^+}$ and $\ce{Cl^-}$ concentrations are equal to the molarity of the acid.

Step 1: Cell Reaction $\ce{\frac{1}{2}H_2(g) + AgCl(s) \to Ag(s) + H^+(aq) + Cl^-(aq)}$ $n = 1$. Quotient $Q = [\ce{H^+}][\ce{Cl^-}] = (10^{-6})(10^{-6}) = 10^{-12}$.

Step 2: Apply Nernst Equation $E_{ \text{cell}} = 0.92 \text{ V}$. $E^\circ_{ \text{cell}} = E^\circ_{\ce{Ag/AgCl/Cl^-}} - 0$. $0.92 = E^\circ_{\ce{AgCl/Ag}} - 0.06 log(10^{-12})$ $0.92 = E^\circ_{\ce{AgCl/Ag}} - 0.06 \times (-12) = E^\circ_{\ce{AgCl/Ag}} + 0.72$ $E^\circ_{\ce{AgCl/Ag}} = 0.92 - 0.72 = 0.20 \text{ V}$

$\boxed{\text{Answer: (b) 0.20 V}}$