Chapter 23 Electrochemistry

Sections 23.1-23.2 Electrochemical Cells OBJECTIVES: Describe how RedOx rxns produce useful electricity Explain the structure and function of Voltaic (Galvanic) Cells [i.e. batteries]

The Nature of Voltaic (Galvanic) Cells You have already seen that when a strip of metal is placed in a solution containing a less active metal, a single replacement rxn will occur This is a classic example of a RedOx rxn. Whether the process is spontaneous or not can easily be predicted by using Table J in your Reference Tables for Physical Setting / CHEMISTRY

Voltaic/Galvanic Cells If the half-rxns which define a RedOx rxn are allowed to occur in separate beakers, the electrons can be made to flow through an external wire and used to perform work The problem is that as the RedOx rxn tries to proceed there is an imbalance of ions in the beakers. Nature will not allow this, so we must supply ions from an external source to keep each solution electrically neutral. The external source of ions is called a “Salt Bridge”. It is composed of a salt- saturated gel contained by a glass U-tube.

The Daniell Cell The Daniell Cell was the first “wet cell” battery. It is composed of a copper electrode in a copper (II) sulfate solution, a zinc electrode in a zinc sulfate solution, a salt bridge (usually containing sodium chloride), an external wire, and a voltmeter. The electrons spontaneously flow from the Zn electrode to the Cu electrode. How the cell works is described in the next slide.

Daniell Cell Half-Rxns Oxidation Half-Rxn: Zn(s) → Zn+2(aq) + 2e-1 ANODE (-) Reduction Half-Rxn: Cu+2(aq) + 2e-1 → Cu(s) CATHODE (+) An Ox, Red Cat

So What Else Happens in the Daniell Cell? The electrons flow from the Zn electrode to the Cu electrode The two RedOx rxns happen simultaneously The cation in the salt bridge moves toward the cathode (Cu Electrode) The anion in the salt bridge moves toward the anode (Zn Electrode) The Zn electrode gets lighter in mass The Cu electrode gets heavier in mass The rxn stops (the battery is dead) when the salt in the salt bridge runs out, or the Zn electrode is used up, or the Cu+2 ions run out

Cell Voltages Standard cell conditions are defined as: 1 Molar Solute, 25◦ C, 1 Atm A “Standard Hydrogen Electrode” under standard conditions has a back voltage applied so the observed cell voltage appears to be zero (see diagrams in the next slides) If a stated voltage (E) has a zero superscript (E◦), the experiment was done under standard conditions

(a) The Standard Hydrogen Cell and (b) Close up of the Hydrogen Standard Electrode.

Cell Voltages You text book has a table of Reduction Potentials (voltages) on pg. 688 Find the two half reactions for your cell; in this case we have: Cu+2(aq) + 2e-1 → Cu(s) E◦ = +0.34 V Zn+2 (aq) + 2e-1 → Zn (s) E◦ = -0.76 V Since you must have both a Red and an Ox rxn, turn the half rxn with the smaller voltage around and change the sign of the voltage (next slide)

Cell Voltages (continued) In the present case we have: Cu+2(aq) + 2e-1 → Cu(s) E◦ = +0.34 V Zn(s) → Zn+2 (aq) + 2e-1 E◦ = +0.76 V If the number of electrons on each side is the same simply add the half rxns together and simplify; the voltages are also added together in a similar fashion: Cu+2(aq) + Zn(s) → Zn+2 (aq) + Cu(s) Overall Rxn +0.34 + (+0.76) = +1.10 Volts E◦cell

The Daniell Cell under Standard Conditions (notice the cell voltage!)

Another Example (pg. 1 of 5) PROBLEM: Suppose someone gave you Al(s), Zn(s), Al(NO3)3(aq), Zn(NO3)2(aq), and NaC2H3O2 [as a paste]… construct a Voltaic cell and label or explain all components SOLUTION: First, identify the electrodes (usually solids), then immerse them in their appropriate electrolytes, and let the paste be the salt bridge… so in this case we have – Al(s) in Al(NO3)3(aq) Zn(s), in Zn(NO3)2(aq) NaC2H3O2 [as a paste] is in the salt bridge

Another Example (pg. 2 of 5) Now write the Reduction Half Rxns using pg. 688: Al+3(aq) + 3e-1 → Al(s) E◦ = -1.66 V Zn+2(aq) + 2e-1 → Zn(s) E◦ = -0.76 V You must have the same number of electrons on both sides of the arrow, so multiple the first rxn by 2 and the second rxn by 3. The voltages, however, are not changed: 2Al+3(aq) + 6e-1 → 2Al(s) E◦ = -1.66 V 3Zn+2(aq) + 6e-1 → 3Zn(s) E◦ = -0.76 V

Another Example (pg. 3 of 5) Turn the smaller voltage rxn around, change sign, and add: 2Al(s) → 2Al+3(aq) + 6e-1 (Ox) E◦ = +1.66 V 3Zn+2(aq) + 6e-1 → 3Zn(s) (Red) E◦ = -0.76 V 2Al(s) + 3Zn+2(aq) → 3Zn(s) + 2Al+3(aq) Over All Net Ionic Rxn (balanced by charge & mass!) E◦cell = +0.90 V = (+1.66 V) + (-0.76 V) Note that all cell voltages must be positive!

Another Example (pg. 4 of 5) Thus, we know: Electrons flow from Al(s) → Zn(s) The cell voltage is +0.90 V Reduction occurs at the Zn electrode, and it is the cathode (+) Oxidation occurs at the Al electrode, and it is the anode (-) Na+1(aq) from the salt bridge flows to the cathode (Zn) C2H3O2(aq)-1 from the salt bridge flows to the anode (Al) The Zn electrode gets heavier while the Al electrode gets lighter

Another Example (pg. 5 of 5)

Other Batteries : Lead Storage Cell

Other Batteries : Standard Dry Cell

Other Batteries : Alkaline Battery

Section 23.3 Electrolytic Cells & More OBJECTIVES: Describe how RedOx rxns can be used to electroplate Discuss some other practical applications of electrochemistry

Electrolytic Cell Basics Electrolytic Cells require an external D.C. power supply The power supply forces a RedOx rxn to take place backwards (rxn normally has a positive DG◦ or a negative E◦) Usually, there is only one beaker containing both the electrolyte and electrode AnOx & RedCat still works .. but the cathode is now (-) and the anode is now (+). This is backwards compared to a Voltaic Cell!

Electroplating of Silver Metal The oxidation of sliver metal has an E◦ value of -0.80 V, so it is not a spontaneous reaction. The external power source supplies this voltage to drive the rxn as seen in the adjacent diagram. The silver electrode is oxidized (loses weight) and the spoon is plated by the reduction of silver ions.

The Formation of Rust Water acts as the medium for the RedOx rxn between solid Fe (oxidized) and molecular oxygen (reduced) + water. The rxn causes a pit to form that will eventually go all the way though the metal. Some metals, such as Al or Ag, form a protective oxidized layer that prevents pitting.

How to Prevent Rust Coat the metal with paint or lacquer to seal out oxygen Use a “sacrifice metal” This implies a metal that oxidizes (rusts) easier than the metal you which to protect, is allowed to rust such that the electrons it produces are fed into the protected metal (see next slide) Apply an external D.C. voltage This is essentially the same as the previous method, but instead of a sacrifice metal a battery is used

One Type of Cathodic Protection

Tarnishing Some metals, such as aluminum and silver, form a thin layer of oxide (rust), which protects the metal from corroding or pitting all the way through Other metals, such as iron, have no such protection and rusts all the way through to the other side Sorry… this is the last time we will use the “slide notes” this year!!!