Question 21.2: Describing a Voltaic Cell with a Diagram and Notation Proble......

Plan From the given contents of the half-cells, we write the half-reactions. To determine which is the anode compartment (oxidation) and which is the cathode (reduction), we note the relative electrode charges (which are based on the direction of the spontaneous redox reaction). Electrons are released into the anode during oxidation, so it has a negative charge. We are told that Cr is negative, so it is the anode and Ag is the cathode.

Solution Writing the balanced half-reactions. The Ag half-reaction consumes e^−:
              Ag^+(aq)  +  e^−  ⟶  Ag(s)          [reduction; cathode]
The Cr half-reaction releases e^−:
              Cr(s)  ⟶  Cr^{3+}(aq)  +  3e^−           [oxidation; anode]
Writing the balanced overall cell equation. We triple the reduction half-reaction to balance e^− and combine the half-reactions:
              Cr(s)  +  3Ag^+(aq)  ⟶  Cr^{3+}(aq)  +  3Ag(s)

Determining direction of electron and ion flow. The released e^− in the Cr electrode (negative) flow through the external circuit to the Ag electrode (positive). As Cr^{3+} ions enter the anode electrolyte, NO_3^− ions enter from the salt bridge to maintain neutrality. As Ag^+ ions leave the cathode electrolyte and plate out on the Ag electrode, K^+ ions enter from the salt bridge to maintain neutrality. The cell diagram is shown in the margin (Fig 21.2). Writing the cell notation with the oxidation components on the left and the reduction components on the right:

Check Always be sure that the half-reactions and the cell reaction are balanced, the half-cells contain all components of the half-reactions, and the electron and ion flow are shown. You should be able to write the half-reactions from the cell notation as a check.
Comment The diagram of a voltaic cell relies on the direction of the spontaneous reaction to give the oxidation (anode; negative) and reduction (cathode; positive) half reactions.