19.1 ELECTROCHEMICAL CELLS

ESSENTIAL IDEA Energy conversions between electrical and chemical energy lie at the core of electrochemical cells. NATURE OF SCIENCE (3.1) Employing quantitative reasoning – electrode potentials and the standard hydrogen electrode. NATURE OF SCIENCE (4.5) Collaboration and ethical implications – scientists have collaborated to work on electrochemical cell technologies and have to consider the environmental and ethical implications of using fuel cells and microbial fuel cells.

INTERNATIONAL-MINDEDNESS Many electrochemical cells can act as energy sources alleviating the world’s energy problems but some cells such as super-efficient microbial fuel cells (MFCs) (also termed biological fuel cells) can contribute to clean up of the environment. How do national governments and the international community decide on research priorities for funding purposes?

THEORY OF KNOWLEDGE The SHE (standard hydrogen electrode) is an example of an arbitrary reference. Would our scientific knowledge be the same if we chose different references?

UNDERSTANDING/KEY IDEA 19.1.A A voltaic cell generates an electromotive force (EMF) resulting in the movement of electrons from the anode (negative electrode) to the cathode (positive electrode) via the external circuit. The EMF is termed the cell potential (E ◦ ).

STANDARD ELECTRODE POTENTIALS  Voltaic cells generate an electromotive (emf) measured in volts as electrons flow from the half-cell with the more negative potential to the half-cell with the more positive potential.  The magnitude of this voltage depends upon the difference in the tendencies of the 2 half- cells to undergo reduction.  No one half-cell can be measured in isolation.  We need a fixed reference point that acts as a standard for measurement – Standard Hydrogen Electrode.

UNDERSTANDING/KEY IDEA 19.1.B The standard hydrogen electrode (SHE) consists of an inert platinum electrode in contact with 1 mol/dm 3 hydrogen ion and hydrogen gas at 100 kPa and 298 K. The standard electrode potential (E ◦ ) is the potential (voltage) of the reduction half- equation under standard conditions measured relative to the SHE. Solute concentration is 1 mol/dm 3 or 100kPa for gases. (E ◦ ) of the SHE is 0V.

STANDARD HYDROGEN ELECTRODE  The standard hydrogen electrode is assigned an electrode potential of 0V.  The concentration of the HCl is 1.0 mol/dm 3.  The H 2 gas is at 298K and 100 kPa.  The metal is platinum.

 Platinum is chosen because it is a fairly inert metal that will not ionize.  The reaction on the electrode happens rapidly as the large surface area helps with the adsorption of hydrogen gas.  Adsorption only occurs on the surface.  As the electrode is immersed in the acid, an equilibrium is set up between the adsorbed layer of H 2 gas and the H + ions. 2H + (aq) + 2e - → H 2(g)

 The reaction is reversible, occuring as the reduction of H + (forward rxn) or the oxidation of H 2 (backward rxn) depending upon the electrode potential of the half-cell to which it is linked.  The hydrogen half-cell is arbitrarily assigned an electrode potential of 0V. This gives us a means to measure and compare the electrode potential of any other half-cell to which it is connected.

STANDARD CONDITIONS  The following variables must be controlled to compare standard electrode potentials: All solutions must be 1.0 mol/dm 3 All gases must have a pressure of 100kPa All substances must be pure. Temperature is 298K or 25˚C. If the half-cell does not include a metal, platinum must be used as the electrode.

STANDARD ELECTRODE POTENTIAL E ˚  Half-cells under the conditions listed previously are known as standard half-cells.  When the standard hydrogen electrode is connected to another standard half-cell, the emf generated is known as the standard electrode potential (E˚) of that half-cell.  E means electrode potential and ˚ means carried out at standard conditions.

 Positive values for E˚ mean that electrons flow from hydrogen to the metal electrode. The hydrogen is oxidized (anode) and the metal is reduced (cathode).  Negative values for E˚ mean that electrons flow from the metal to the hydrogen electrode. The hydrogen is reduced (cathode) and the metal is oxidized (anode).

 Let’s refer back to our previous activity series to make sense of all of this. Mg V strongest reducing agent Al V (most readily oxidized) Zn V Fe V Pb V Hydrogen falls here 0.00 V Cu Ag +0.80weakest reducing agent (least readily oxidized) Notice that H 2 would be the cathode for the metals above it and the anode for the metals below it on the series.

 The more negative the E˚ the stronger the reducing agent and the more readily it loses electrons or becomes oxidized.  The more positive the E˚ the more it tends to be reduced.  In a voltaic or galvanic cell, electrons always flow from the half-cell with the more negative electrode potential to the half-cell with the more positive electrode potential.  The more negative (lower) E˚ is always the anode and the more positive (higher) E˚ is always the cathode.

APPLICATION/SKILLS Be able to calculate cell potentials using standard electrode potentials.

CALCULATING CELL POTENTIALS  Electrons always flow towards the half-cell with the highest or most positive E˚.  The standard electrode potentials are listed in your IB Data booklet.  All values listed are reduction potentials so you will have to reverse one of the values for the species being oxidized.  For a galvanic or voltaic cell, E˚ will always be positive indicating a spontaneous reaction.

 Oxidation occurs with the E˚ of the more negative or lower value.  Reduction occurs with the E˚ of the more positive or higher value.  Do not multiply the E˚ by a factor as you do in Hess’s law even if you had to multiply the equations by a factor to get the electrons to cancel out.

USES OF THE E ˚ DATA  You can predict spontaneity by using the reaction as it is given and calculating E˚.  You can also calculate the emf of a voltaic cell.  You can be given 2 half reactions and be expected to draw the cell, write the overall reaction and determine E˚ cell.  You can compare oxidizing and reducing power of half cells. (The more positive or higher E˚ is the strongest oxidizing agent.)

APPLICATION/SKILLS Be able to predict whether a reaction is spontaneous or not using electrode potential (E ◦ ) values.

 If your calculations give you a positive E˚ cell, then the reaction is spontaneous.  Remember that even though the E˚ can give you the feasibility of a reaction, it does not give you any information on the rate of the reaction.  The activation energy may be too high for the reaction to occur at any appreciable rate.  Remember this on fundamental fact: Electrons ALWAYS flow to the half-cell with the highest or most positive value for E˚.

UNDERSTANDING/KEY IDEA 19.1.C ΔG ◦ = -nFE ◦. When E is positive, ΔG ◦ is negative indicative of a spontaneous process. When E ◦ is negative, ΔG ◦ is positive indicative of a non-spontaneous process. When E ◦ is 0, ΔG ◦ is 0.

APPLICATION/SKILLS Be able to determine the standard free-energy changes ΔG ◦ using standard electrode potentials.

 We now have 2 ways to measure the spontaneity of a reaction, E ◦ and ΔG ◦  They are related by the equation ΔG ◦ = -nFE ◦ Where n is the number of moles of electrons. F (Faraday’s constant) is the charge carried by 1 mol of electrons = 96,500 C/mol

 This equation allows us to calculate the free-energy change of a reaction from standard electrode potential data.  The more positive the value for E, the more favorable the reaction or more negative the delta G.  A voltmeter is an indirect measure of free- energy change as well as electrode potential.  See sample problem on page 400.

UNDERSTANDING/KEY IDEA 19.1.D When aqueous solutions are electrolyzed, water can be oxidized to oxygen at the anode and reduced to hydrogen at the cathode.

APPLICATION/SKILLS Be able to explain the products formed during electrolysis of aqueous solutions.

 The electrolysis of an aqueous solution involves ions from the solvent water as well as the ions from the ionic compound.  More than one type of ion then migrates to the anode and the cathode.  Selective discharge is a process by which certain ions are discharged at each electrode.

 The products of electrolysis are determined by the following factors that influence the discharge of ions: The relative E˚ of the ions The relative concentrations of the ions in the electrolyte The nature of the electrode.

ELECTROLYSIS OF WATER  The ionization of pure water is very low.  However, the addition of ions increases its conductivity so some ionic compound like NaOH is added when electrolysis is performed.  1. Ions present: H 2 O (l) ↔ H + + OH - NaOH (aq) → Na + + OH - cathode anode

 2. At the cathode: Na + and H + accumulate. E˚ for Na + is -2.71V and E˚ for H + is 0.00V H + is preferentially reduced and H 2 is discharged. 2H + (aq) + 2e - → H 2(g) At the Anode: OH - is discharged as the only ion 4OH - (aq) → 2H 2 O (l) + O 2(g) + 4e -

 3. The overall equation: 2H 2 O (l) → 2H 2(g) + O 2(g)  4. The observed changes at electrodes: A colorless gas evolved at both electrodes – ○ O 2 at anode and H 2 at cathode The ratio of volume of gases is 2H 2 to 1O 2. The pH at the anode decreases as OH - is discharged (taken away from the solution). The pH at the cathode increases as H + is discharged (taken away from the solution).

ELECTROLYSIS OF NaCl (aq)  NaCl (aq) is known as brine.  The electrolysis of brine leads to the production of H 2, Cl 2 and NaOH.  The electrolysis of NaCl (aq) and NaCl (l) are different processes leading to different products.  1. Ions present: NaCl (aq) → Na + + Cl - H 2 O (l) ↔ H + + OH - cathode anode

 2. At the cathode: Na + and H + accumulate. E˚ for Na + is -2.71V and E˚ for H + is 0.00V H + is preferentially reduced and H 2 is discharged. 2H + (aq) + 2e - → H 2(g) At the Anode: Cl - and OH - accumulate. The E˚ for OH - is lower than Cl - so we would expect OH - to be preferentially oxidized, but it is more complicated. ○ When the concentration of Cl- is low, OH- is discharged leading to the release of O 2. ○ But when the concentration of NaCl is greater than 25% by mass of the solution, Cl - is discharged leading to the release of Cl 2. 2Cl - → Cl 2 + 2e - ○ The industrial electrolysis of brine uses saturated solutions of aqueous NaCl to produce Cl 2.

 3. The overall equation when Cl- is discharged: 2NaCl (aq) + 2H 2 O (l) → 2H 2(g) + Cl 2(g) + 2Na + + 2OH -  4. The observed changes at electrodes: A gas evolves at both electrodes – ○ Cl 2 at anode and H 2 at cathode Cl 2(g) is identified at the anode through its strong smell and bleaching effect on blue litmus paper. An increase in pH of the electrolyte occurs due to the loss of H +.

ELECTROLYSIS OF CuSO 4(aq)  CuSO 4(aq) is bright blue due to the hydrated Cu 2+ ion.  The electrolysis leads to different products depending upon the nature of the electrodes: either carbon or copper.  1. Ions present: CuSO 4 (aq) → Cu 2+ + SO 4 2- H 2 O (l) ↔ H + + OH - cathode anode

(A) With Carbon (graphite) as the electrodes 2. At the cathode: Cu 2+ and H + accumulate. E˚ for Cu 2+ is +0.34V and E˚ for H + is 0.00V Cu 2+ is preferentially reduced and Cu (s) is discharged. Cu 2+ (aq) + 2e - → Cu (s) At the Anode: SO 4 2- and OH - accumulate. The E˚ for OH - is lower than SO 4 2- so OH - is discharged. 4OH - (aq) → 2H 2 O (l) + O 2(g) + 4e -

 3. The overall equation is: 2 Cu 2+ (aq) + 2H 2 O (l) → 2Cu (s) + O 2(g) + 4H +  4. The observed changes at electrodes: Pinky-brown color develops as copper is deposited on the cathode. A colorless gas O 2 is evolved at the anode. A decrease in pH of the electrolyte occurs due to the loss of OH -. Loss of intensity of blue color as Cu 2+ is discharged.

(B) With copper as the electrodes 2. At the cathode: Cu 2+ and H + accumulate. E˚ for Cu 2+ is +0.34V and E˚ for H + is 0.00V Cu 2+ is preferentially reduced and Cu (s) is discharged. Cu 2+ (aq) + 2e - → Cu (s) At the Anode: The Cu electrode itself is oxidized, supplying electrons for the reaction and dissolving as Cu 2+. Cu (s) → Cu 2+ (aq) + 2e -

 3. So the net reaction is the movement of Cu 2+ from where it is produced at the anode to the cathode where it is discharged as Cu (s).  4. The observed changes at electrodes: Pinky-brown color develops as copper is deposited on the cathode. Disintegration of the Cu anode. No change in pH. No change in intensity of blue color as the concentration of Cu 2+ remains constant.

UNDERSTANDING/KEY IDEA 19.1.E Current, duration of electrolysis and charge on the ion affect the amount of product formed at the electrodes during electrolysis.

APPLICATION/SKILLS Be able to determine the relative amounts of products formed during electrolytic processes.

 The three factors influencing the amount of products are: The charge on the ion The current The duration of the electrolysis  There are two types of problems involving electrolysis.

STEPS TO SOLVE FOR AMOUNT OF PRODUCT IN ELECTROLYSIS PROBLEMS  Step 1: Multiply your current by your time. 1 ampere (amp) = 1 C/s where C = coulomb and time needs to be in seconds.  Step 2: Multiply coulombs by mole e - /96500 C to get moles of electrons  Step 3: Divide by moles of electrons in the process.  Step 4: Multiply by molar mass to get grams.

GUIDANCE Faraday’s constant = 96,500 C/mol is found in the data booklet.

EXAMPLE  How much copper will be plated out when a current of 10.0 amps is passed through a solution of Cu 2+ ions for 30.0 minutes? Cu e - → Cu(s)  Step 1: Multiply your amps by your time (must convert minutes to seconds first) 10.0 amps = 10.0 C/s x 30.0 min x 60 sec/min = 1.80 x 10 4 C  Step 2: Multiply coulombs by 1 mol e - /96500 C to get moles of e x 10 4 C x 1mol e - /96500 C = 1.87 x mol e-  Step 3: Divide by moles of electrons in equation..187 mol e - x 1 mol Cu/2 mol e - =.0935 mol Cu  Step 4: Multiply by molar mass to get grams.0935 mol Cu x 63.55g/mol = 5.94 g Cu

SOLVING FOR TIME IN ELECTROLYSIS  Step 1: Divide grams by molar mass to get moles.  Step 2: Multiply by moles of electrons.  Step 3: Multiply moles of electrons by C/moles e -  Step 4: Divide by amps and convert to minutes.

 How long must a current of 5.00 A be applied to a solution of Ag + to produce 10.5 g of silver?  Step 1: Divide by molar mass to get moles g x mol/107.9g =.0973 mol  Step 2: Multiply by moles of electrons mol Ag x 1 mol e-/1 mol Ag =.0973 mol e-  Step 3: Multiply by Faraday’s constant to get Coulombs.0973 mol e- x C/mol e- = 9390 C  Step 4: Divide by the given amps and convert to minutes 9390 C x s/5.00C x min/60s = 31.3 min

UNDERSTANDING/KEY IDEA 19.1.F Electroplating involves the electrolytic coating of an object with a metallic thin layer.

APPLICATION/SKILLS Be able to explain the process of electroplating.

ELECTROPLATING  Electroplating is the process of using electrolysis to deposit a layer of metal on top of another metal or conductive substance.  It must have the following features: An electrolyte containing the metal ions which are to be deposited The cathode made of the object to be plated The anode can be the same metal if it is used to replenish the supply of ions in the electrolyte. Process can be controlled by altering the current and time according to how thick a layer of metal is desired.

PURPOSES OF ELECTROPLATING  Decorative purposes  Corrosion control  Improvement of function

Citations International Baccalaureate Organization. Chemistry Guide, First assessment Updated Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed. N.p.: Pearson Baccalaureate, Print. Most of the information found in this power point comes directly from this textbook. The power point has been made to directly complement the Higher Level Chemistry textbook by Catrin and Brown and is used for direct instructional purposes only.