Question 25.2: Consider the electrochemical cell shown in Figure 25.7. Indi......
Consider the electrochemical cell shown in Figure 25.7. Indicate the current flow in the external circuit and in the cell electrolyte solutions.
From the figure we see that the Pb(s) electrode is the negative electrode and the Ag(s) electrode is the positive electrode. Therefore, electrons flow in the external circuit from the Pb(s) electrode to the Ag(s) electrode. The electrons are produced at the Pb(s) electrode according to
Pb(s) → Pb^{2+}(aq) + 2\ e^-and are consumed at the Ag(s) electrode according to
2\ Ag^+(aq) + 2\ e^− → 2\ Ag(s)Note that two silver ions are reduced for each lead atom oxidized in order to maintain electrical neutrality.
Positive Pb^{2+}(aq) ions are produced at the lead electrode. For each Pb^{2+}(aq) ion produced, two NO_{3}^-(aq) ions flow from the salt bridge into the Pb(NO_{3})_{2}(aq) solution to maintain electrical neutrality. Two positive Ag^+(aq) ions are consumed at the silver electrode for each Pb^{2+}(aq) ion produced; thus, two K^{+}(aq) ions flow simultaneously from the salt bridge into the AgNO_{3}(aq) solution. This flow of ions maintains electrical neutrality both in the two cell solutions and in the salt bridge solution. The net ionic equation that describes the overall cell reaction is
Pb(s) + 2\ Ag^+(aq) → Pb^{2+}(aq) + 2\ Ag(s)If we simply place a lead rod in AgNO_{3}(aq), then the spontaneous reduction of Ag^+(aq) by Pb(s) occurs directly on the surface of the lead, where silver metal crystals form. Separating the two half reactions in an electrochemical cell forces the reaction to proceed through the external circuit and thereby prevents the direct reaction of lead metal with silver ions.