TOPIC C DRIVING FORCES, ENERGY CHANGES, AND ELECTROCHEMISTRY

Driving forces Evidence for chemical change can manifest itself in a number of ways. The formation of precipitate a change of energy in the form of heat or light a color change the formation of a gas are all observations that can be made in the laboratory Sometimes these events are called driving forces, but are they chemical or physical? an interruption in the inter-molecular forces ….. change is physical a re-arrangement of the intra-molecular forces ….. change is chemical.

Practice: 1. Discuss the change in forces and bonds when water boils. 2. Discuss the change in forces and bonds when water decomposes into its elements.

Gas producing reactions Several reactions produce gases as one of the products. These are worth learning, as well as the subsequent tests for the gases produced. General gas producing reactions 1. Acid + Metal  Salt + Hydrogen For example: Zn(s) + H 2 SO 4 (aq)  ZnSO 4 (aq) + H 2 (g) Test for gas: “Squeaky pop” with lighted splint

2. Acid + Carbonate  Salt + Water + CO 2 For example: H 2 SO 4 (aq) + CaCO 3 (s)  CaSO 4 (aq) + H 2 O(l) + CO 2 (g) Test for gas: Extinguishes a glowing or lighted splint Turns lime water (Ca(OH) 2 ) milky Specific gas producing reaction 1. The production of O 2 by the decomposition of H 2 O 2 (with MnO 2 catalyst) 2H 2 O 2 (aq)  2H 2 O(l) + O 2 (g) Test for gas: Relights glowing splint.

Energy Changes When reactants undergo a chemical change – products are formed. The reaction is considered a part of the universe called the system Everything else is called the surroundings Energy of Universe = Energy of System + Energy of Surroundings The system will often undergo an energy change where it will either: release energy to the surroundings (exothermic reaction) absorb energy from the surroundings (endothermic reaction)

For an exothermic reaction - temperature of the surroundings increases For an endothermic reaction - temperature of the surrounding decreases These changes can be shown graphically in energy diagrams

Electrochemistry When a metal comes into contact with a solution containing its own ions an equilibrium is set up. M x+ (aq) + xe-   M(s) Some reactive metals (like Mg) will lose electrons readily The equilibrium lies to the left. A large number of electrons are released which collect on the surface of the metal giving a negative charge Mg 2+ (aq) + 2e-   Mg(s)

Less reactive metals (Ag) show less tendency to ionize equilibrium lies to the right Fewer electrons will collect on the metal and the charge will be much less negative In fact, if the aqueous ions remove electrons from the metal it will develop a positive charge. Ag+(aq) + e-   Ag(s)

Non-metals can also be considered, for example: H + (aq) + e-   ½H 2 (g) When an element is placed in a solution containing its own ions - an electric charge will develop on the metal (In the case of a non-metal – an inert conductor is used) The charge is called the electrode potential The system is called a half-cell. The sign and size of the charge will depend on the ability of the element to lose or gain electrons.

Electrochemical series and electrode potentials Species that appear at the top of the series: gain electrons most readily have the most positive E o values are easily reduced and (best oxidizing agents) Species that appear at the bottom of the series: lose electrons most readily have the most negative E o values are easily oxidized and (best reducing agents)

Electrochemical cells (batteries) An electrochemical cell Generates electrical energy from a spontaneous Redox reaction. Connecting two half-cells that have different electrode potentials forms an electrochemical cell (battery). A voltmeter is used to measure the voltage A salt bridge connects the two half-cells.

H + MnO 4 - Fe +2 Connected this way the reaction starts Stops immediately because charge builds up. MnO e-  MnO 4 2- E o = 0.56 Fe e-  FeE o = -0.44

H + MnO 4 - Fe +2 Galvanic Cell - uses a spontaneous redox reaction to produce a current that can be used to do work. Salt Bridge allows current to flow

H + MnO 4 - Fe +2 e-e- Electricity travels in a complete circuit Oxidation occurs at the anode Reduction occurs at the cathode Anode Cathode

H + MnO 4 - Fe +2 Porous Disk

Reducing Agent Oxidizing Agent e-e- e-e- e-e- e-e- e-e- e-e- AnodeCathode Overview in General

Cell Potential Oxidizing agent pushes the electron. Reducing agent pulls the electron. The push or pull (“driving force”) is called the cell potential E cell Also called the electromotive force (emf) Unit is the volt(V) = 1 J/C (joule of work/coulomb of charge) measured with a voltmeter A coulomb is the SI unit of quantity of electricity, (the charge transferred in one second with a constant current of one ampere)

Zn +2 SO M HCl Anode M ZnSO 4 H + Cl - H 2 in Cathode Zn metal e-e- Pt metal

1 M HCl H + Cl - H 2 in Standard Hydrogen Electrode This is the reference all other oxidations are compared to Eº = 0 º indicates standard states of 25ºC, 1 atm, 1 M solutions.

Standard Reduction Potentials It is universally accepted that the half-reaction potential for 2H + + 2e- → H 2 assigned a value of zero volts. The overall cell potential (E cell ) is then assigned to the other half reaction. This reaction is always written as a reduction potential Table 17.1 has a list of the most common Standard Reduction Potentials we will use.

Finding Cell Potentials using Standard Reduction Potentials Fe 3+ (aq) + Cu(s)  Cu +2 (aq) + Fe 2+ (aq) Two half reactions:  Fe 3+ + e- → Fe 2+ E o = 0.77 V  Cu e- → Cu E o = 0.34 V Two rules apply:  1. The half reaction with the largest potential is written as a reduction, the other must be reversed (change its sign).  2. The number of electrons lost must equal the number of electrons gained (multiply half reactions).  Do not multiply the standard potentials! They do not change!

The total cell potential is the sum of the potential at each electrode. E o cell = E o (cathode) + E o (anode) E o cell = 0.77 V + (– 0.34 V) = 0.43 V We can look up reduction potentials in table 17.1.

Line Notation solid  Aqueous  Aqueous  solid Anode on the left  Cathode on the right Single line different phases. Double line porous disk or salt bridge. If all the substances on one side are aqueous, a platinum electrode is indicated. For the last reaction Cu(s)  Cu +2 (aq)  Fe +2 (aq),Fe +3 (aq)  Pt(s)

Galvanic Cell  The reaction always runs spontaneously in the direction that produced a positive cell potential.  Four things for a complete description. 1)Cell Potential 2)Direction of flow 3)Designation of anode and cathode 4)Nature of all the components- electrodes and ions

Practice: Using the SERP table, write cell diagrams for the following combinations of electrodes. Remember to include state symbols and inert electrodes where appropriate (i) Zn/Zn 2+ and fluorine (ii) Sn/Sn 2+ and hydrogen (iii) Cu/Cu 2+ and Fe/Fe 2+ (iv) Fe 2+ /Fe 3+ and hydrogen

Practice: A cell formed from a silver standard electrode and a hydrogen standard electrode generates a voltage of V. Hydrogen is the more negative electrode and has a value of 0.00 V. (a) Write the cell diagram and calculate the standard potential of the silver electrode. (b) When the silver electrode is combined with a standard aluminum electrode the voltage of the cell is V and aluminum is the more negative electrode. Write the cell diagram and calculate the standard electrode potential of the aluminum electrode.

 ΔG = Gibbs Free Energy  – the energy available for the cells to do work As long as the cell has a potential (+E o ) it can do work As the cell runs is approaches equilibrium At Equilibrium, E o = 0, ΔG = 0, and K can be calculated Potential, Work and  G

The Relationship between Gibbs Free Energy and E cell Summarized by the expression: ΔG o = -nFE o Where F = Faraday = 96,485 C / mol (coulombs / mols e-) and n = # of moles of electrons transferred Gibbs free energy can also be expressed as: ΔG o = -RT lnK (R = 8.31 J/K · mol) Setting the two equal to each other and solving for E o E o = (RT / nF) lnK

Potential, Work and  G  Gº = -nFE º If E º 0 nonspontaneous if E º > 0, then  Gº < 0 spontaneous In fact, reverse is spontaneous. Calculate  Gº for the following reaction: Cu +2 (aq)+ Fe(s)  Cu(s)+ Fe +2 (aq) Fe +2 (aq) + e -  Fe(s) E º = 0.44 V Cu +2 (aq)+2e -  Cu(s) E º = 0.34 V

Possible combinations of K, E, and ΔG o leads to the following conclusions: KEoEo ΔG o Conclusion 11PositiveNegativeSpontaneous cell reaction = 100At equilibrium < 1NegativePositiveReaction is spontaneous in the reverse direction

Cell Potential and Concentration Qualitatively - Can predict direction of change in E from LeChâtelier. 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) E o cell = 0.48v Predict if E cell will be greater or less than Eº cell If [Al +3 ] = 1.5 M and [Mn +2 ] = 1.0 M if [Al +3 ] = 1.0 M and [Mn +2 ] = 1.5M if [Al +3 ] = 1.5 M and [Mn +2 ] = 1.5 M

 G =  Gº +RTln(Q) -nFE E = -nFEº + RTln(Q) E = E º - RT ln(Q)  This is not commonly used ! nF 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) E º = 0.48 V Consider a cell at 25 o C where:  [Mn 2+ ] = 0.50 M and [Al 3+ ] = 1.5 M Use the Nernst equation on the next slide to solve. The Nernst Equation

E = E º log(Q) used at 25 o C n As reactions proceed concentrations of products increase and reactants decrease. The cell will discharge until it reaches equilibrium. At this point: Q = K (the equilibrium constant) and E cell = 0 At equilibrium, the components in the two cells have the same free energy and ∆G=0. The Cell no longer has the ability to do work.

Practice: If the reaction Zn(s) + Cu 2+ (aq)  Cu(s) + Zn 2+ (aq) is carried out using solutions that are 5.0M Zn 2+ and 0.3M Cu 2+ at 298 K, predict the effect on the voltage of the cell, when compared to the voltage generated under standard conditions.

Electrolysis Running a galvanic cell backwards. Put a voltage bigger than the potential and reverse the direction of the redox reaction. Used for electroplating.

1.0 M Zn +2 e-e- e-e- Anode Cathode 1.10 Zn Cu 1.0 M Cu +2

1.0 M Zn +2 e-e- e-e- Anode Cathode A battery >1.10V Zn Cu 1.0 M Cu +2

Steps: 1. Current and time → charge amps (C/s) x time (s) to get coulombs 2. Quantity of charge → mol e- coulombs (C) x 1/F (mol e- / C) to get mol e- 3. Moles of e- → mol of element mol e- x 1 mol of element / mols of e- (needed to form neutral element from ion) 4. Moles of element → mass of element mole of element x molar mass / 1 mol

Calculating plating Have to count charge. Measure current I (in amperes) 1 amp = 1 coulomb of charge per second q = I x t q/nF = moles of metal Mass of plated metal

 What mass of copper is plated out when a current of 10.0 amps is passed for 30.0 minutes through a solution containing Cu 2+.  How long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag +

Other uses Electroysis of water. Separating mixtures of ions. More positive reduction potential means the reaction proceeds forward. For metals this is typically gaining electrons and forming the solid metal – this removes the ion from solution. Ions with the more positive the reduction potential will “plate out” first.