Inert Electrodes The zinc anode and copper cathode of a Daniell cell and the silver and chromium electrodes in Figure 13.4 are all metals and can act as electrical conductors. However, some redox reactions involve oxidizing and reducing agents that are not solid metals but, instead, are dissolved electrolytes or gases and, therefore, cannot be used as electrodes. To construct a voltaic cell that will use these oxidizing and reducing agents, you have to use inert electrodes. An inert electrode is an electrode made from a material that is neither a reactant nor a product of the redox reaction. Instead, the inert electrode can carry a current and provide a surface on which redox reactions can occur. Figure 13.5 shows a cell that contains one example of an inert electrode—a platinum electrode. The complete balanced equation, net ionic equation, and half-reactions for this cell are given below. complete balanced equation: Pb(s) + 2FeCl3(aq) → 2FeCl2(aq) + PbCl2(aq) net ionic equation: Pb(s) + 2Fe3+(aq) → 2Fe2+(aq) + Pb2+(aq) oxidation half-reaction: Pb(s) → Pb2+(aq) + 2e– reduction half-reaction: Fe3+(aq) + e– → Fe2+(aq) The anode is the lead electrode. Lead atoms lose electrons that remain in the electrode while the lead(II) ions dissolve in the solution in the same way that the anode did in previous example. However, the reduction half-reaction involves iron(III) ions that accept an electron from the platinum inert electrode and become iron(II) ions. The platinum atoms in the electrode (cathode) remain unchanged. Voltmeter e- e- Anode Salt bridge Cathode (-) K+ (+) Pb Cl- Pt 2e- e- Fe3+ Pb Fe2+ Pb2+ PbCl2 FeCl3 FeCl2 Figure 13.5 This cell uses an inert electrode to conduct electrons. Why do you think that platinum is often chosen as an inert electrode? Another common choice is graphite. Cell Notation Sketching entire cells helps you to visualize and learn about the movement of ions and electrons in a voltaic cell. However, it becomes tedious when you are trying to communicate quickly. Chemists have therefore developed a shorthand method for representing voltaic cells called a cell notation. The cell notation of a Daniell cell is as follows: Zn (s) Zn2+( aq) Cu2+( aq) Cu (s) Chapter 13 Cells and Batteries • MHR 481 ChemistryFile In the cell notation, as in the sketch of the entire cell, the anode is usually shown on the left and the cathode on the right. Each single vertical line, , represents a phase boundary FYI The cell notation is correctly between the electrode and the solution in a half-cell. For example, the fi rst single vertical called a cell diagram. line shows that the solid zinc and aqueous zinc ions are in different phases or states— However, some people solid and aqueous. The double vertical line, , represents the porous barrier or salt confuse the term cell bridge between the half-cells. diagram with a drawing of Inert electrodes, such as the platinum electrode in the cell shown in Figure 13.5, do a complete cell. Therefore, this text uses the term not appear in the chemical equation or in half-reactions. However, they are included in cell notation. the cell notation. The cell notation for the cell in Figure 13.5 is shown here. Pb (s) Pb2+( aq) Fe3+( aq) , Fe2+( aq) Pt (s) A comma separates the formulas Fe3+ and Fe2+ ions involved in the reduction half- reaction to indicate that they are in the same phase—the aqueous phase. • • • 4 a) If the reaction of zinc with copper(II) ions is carried out in a test tube, what Q is the oxidizing agent and what is the reducing agent? b) In a Daniell cell, what is the oxidizing agent and what is the reducing agent? Explain your answer. Q5 Sketch voltaic cells for each of the following cell notations. Label all parts of the sketches. Below the sketch, write the oxidation half-reaction, the reduction half-reaction, and the overall cell reaction. Identify the anode and the cathode in each case. Identify any inert electrodes. Explain why there is a line instead of a comma between the symbols for the hydrogen ions and the hydrogen gas in reaction (b). a) Sn (s) Sn2+( aq) T1+( aq) T1 (s) 2+ + b) Cd (s) Cd ( aq) H ( aq) H2( g) Pt (s) Q6 A voltaic cell involves the overall reaction of iodide ions with acidifi ed permanganate ions to form manganese(II) ions and iodine. The salt bridge contains potassium nitrate. Both electrodes are inert and are made of graphite. a) Write the half-reactions and the overall cell reaction. b) Identify the oxidizing agent and the reducing agent. c) Solid iodine forms on one of the electrodes. Does it form on the anode or the cathode? Explain. d) Sketch and label the entire voltaic cell. Q7 As you saw earlier, pushing a zinc electrode and a copper electrode into a lemon makes a “lemon cell.” In the following representation of the cell, C6H8O7 is the formula for citric acid. Explain why the representation does not include a double vertical line: Zn (s) C6H8O7 ( aq) Cu (s) • • • Cell Potentials You read that the oxidation reaction at the anode causes the anode to become negatively charged and the reduction reaction that occurs at the cathode causes the cathode to become positively charged. Whenever a separation of charge exists, a force acts 482 MHR • Unit 6 Electrochemical Changes Table 13.1 Standard Reduction Potentials (298 K, 1 atm) Half-reaction E° (V) - - agent as reducing strength Increasing F2(g) + 2e → 2F (aq) +2.87 - - Br2() + 2e → 2Br (aq) +1.07 Ag+(aq) + e- → Ag(s) +0.80 - - I2(s) + 2e → 2I (aq) +0.54 Cu2+(aq) + 2e- → Cu(s) +0.34 + - 2H (aq) + 2e → H2(g) 0.00 Fe2+(aq) + 2e- → Fe(s) -0.45 Cr3+(aq) + 3e- → Cr(s) -0.74 Zn2+(aq) + 2e- → Zn(s) -0.76 Increasing strength as oxidizing agent as oxidizing strength Increasing Al3+(aq) + 3e- → Al(s) -1.66 Na+(aq) + e- → Na(s) -2.71 If you compare the sequence of the metals in Table 13.1 with the table of oxidizing and reducing agents in Table 12.2, you will discover that the sequences are the same. Table 12.2 showed the relative strengths of the ions and atoms as oxidizing or reducing agents. In this table, however, you have numerical values that tell you how much stronger the atoms, ions, or compounds are as oxidizing or reducing agents. Another important application of these data is in choosing the salts for a salt bridge in a voltaic cell. Re-examine the Daniell cell as it is reproduced in Figure 13.8. Suppose, for example, that you had chosen silver nitrate, AgNO3(aq), for the salt bridge. Silver, rather than potassium, would diffuse into the copper(II) sulfate solution. Silver ions are stronger oxidizing agents than are copper(II) ions, so the silver ions would migrate to the copper anode and accept the electrons in place of copper(II) ions. As a result, the reading on the voltmeter would be incorrect. 6OLTMETER Figure 13.8 When potassium diffuses into E E the copper(II) sulfate solution, it has no effect !NODE 3ALTBRIDGE #ATHODE on the reduction reaction + :N #L #U of copper. However, any ion that has a more positive standard reduction E E potential than does copper(II) will displace :N the copper(II) ions at the :N #U cathode. :N #U Calculating Standard Cell Potentials You can use Table 13.1 to calculate the standard cell potential, from the formula below, of any cell for which the standard reduction potentials of the oxidizing and reducing agents are listed on a table. E °cell = E °cathode – E °anode E °cell: standard cell potential E °cathode: standard reduction potential for the reduction (cathode) half-reaction E °anode: standard reduction potential for the oxidation (anode) half-reaction Chapter 13 Cells and Batteries • MHR 485 Table 13.2 Comparison of Voltaic Cell and Electrolytic Cell ChemistryFile Voltaic cell Electrolytic cell FYI spontaneous reaction non-spontaneous reaction To remember the sign of converts chemical energy to electrical energy converts electrical energy to chemical energy the charge on the cathode in the two different types anode (negatively charged): anode (positively charged): of cells (voltaic cells versus zinc electrode copper electrode electrolytic cells), remember cathode (positively charged): cathode (negatively charged): that electrons always copper electrode zinc electrode leave the cathode and a oxidation (at anode): oxidation (at anode): reduction half-reaction Zn(s) → Zn2+(aq) + 2e− Cu(s) → Cu2+(aq) + 2e− occurs there. In a voltaic cell, the spontaneous reduction (at cathode): reduction (at cathode): chemical reaction “pulls” 2+ − 2+ − Cu (aq) + 2e → Cu(s) Zn (aq) + 2e → Zn(s) the electrons off the cell reaction: cell reaction: cathode leaving it positively Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Cu(s) + Zn2+(aq) → Cu2+(aq) + Zn(s) charged. In an electrolytic cell, the external source of Previously, you calculated the standard cell potential for the redox reaction between zinc electrical energy “pushes” atoms and copper(II) ions and found it to be 1.10 V. electrons onto the cathode making it negatively E°cell = E°cathode - E°anode charged. This negatively E°cell = E°copper - E°zinc charged cathode then “pushes” electrons onto E°cell = +0.34 V - (-0.76 V) the compound in solution, E° = +1.10 V cell causing a reduction half- Similarly, if you calculated the standard cell potential for the reverse reaction, you would reaction to occur. obtain −1.10 V. E°cell = E°cathode - E°anode E°cell = E°zinc - E°copper E°cell = -0.76 V - (+0.34 V) E°cell = -1.10 V The negative sign means that the reaction is not spontaneous.