The half-cell comprising functions correctly as a metal-insoluble salt electrode. Which of the follo — Electrochemistry Chemistry Question

💡 Solution & Explanation

The reduction half-reaction is $AgBr(s) + e^- \rightleftharpoons Ag(s) + Br^-(aq)$. The Nernst equation is $E = E^\circ_{Br^-/AgBr/Ag} - \frac{2.303RT}{F}\log[Br^-]$. By substituting $[Ag^+] = K_{sp}/[Br^-]$ into the standard $Ag^+/Ag$ Nernst equation and equating the two, we derive $E^\circ_{Br^-/AgBr/Ag} = E^\circ_{Ag^+/Ag} + \frac{2.303RT}{F}\log K_{sp}$. The electrode is fundamentally reversible with respect to the $Br^-$ anion. Adding $AgNO_3$ changes $[Ag^+]$, perturbing the $K_{sp}$ equilibrium, altering $[Br^-]$, and thus actively changing the electrode potential.