<Dl> <Dd> In one half - cell, dissolved metal - B cations combine with the free electrons that are available at the interface between the solution and the metal - B electrode; these cations are thereby neutralized, causing them to precipitate from solution as deposits on the metal - B electrode, a process known as plating . </Dd> <Dd> This reduction reaction causes the free electrons throughout the metal - B electrode, the wire, and the metal - A electrode to be pulled into the metal - B electrode . Consequently, electrons are wrestled away from some of the atoms of the metal - A electrode, as though the metal - B cations were reacting directly with them; those metal - A atoms become cations that dissolve into the surrounding solution . </Dd> <Dd> As this reaction continues, the half - cell with the metal - A electrode develops a positively charged solution (because the metal - A cations dissolve into it), while the other half - cell develops a negatively charged solution (because the metal - B cations precipitate out of it, leaving behind the anions); unabated, this imbalance in charge would stop the reaction . </Dd> </Dl> <Dd> In one half - cell, dissolved metal - B cations combine with the free electrons that are available at the interface between the solution and the metal - B electrode; these cations are thereby neutralized, causing them to precipitate from solution as deposits on the metal - B electrode, a process known as plating . </Dd> <Dd> This reduction reaction causes the free electrons throughout the metal - B electrode, the wire, and the metal - A electrode to be pulled into the metal - B electrode . Consequently, electrons are wrestled away from some of the atoms of the metal - A electrode, as though the metal - B cations were reacting directly with them; those metal - A atoms become cations that dissolve into the surrounding solution . </Dd> <Dd> As this reaction continues, the half - cell with the metal - A electrode develops a positively charged solution (because the metal - A cations dissolve into it), while the other half - cell develops a negatively charged solution (because the metal - B cations precipitate out of it, leaving behind the anions); unabated, this imbalance in charge would stop the reaction . </Dd>

Where do cations flow in a galvanic cell
find me the text answering this question