CHEM Pharmacy Week 9: Galvanic Cells Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone:

Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, John Wiley & Sons Australia, Ltd ISBN:

Lecture 25-3 Electrochemistry Blackman, Bottle, Schmid, Mocerino & Wille: Chapter 12, Sections 4.8 and 4.9 Key chemical concepts: Redox and half reactions Cell potential Voltaic and electrolytic cells Concentration cells Key Calculations: Calculating cell potential Calculating amount of product for given current Using the Nernst equation for concentration cells

Lecture 25-4 The measured voltage across the cell is called cell potential (E cell ). This driving force for the reaction is also called ELECTROMOTIVE FORCE (emf) of the cell. Every galvanic cell has a different cell potential. How can we measure the cell potential relative to each species? How can we tabulate cell potentials? Electromotive Force Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (s)

Lecture 25-5 The electromotive force cannot be measured absolutely for one element (we cannot measure E Zn and E Cu in isolation). Only differences in potential have meaning (ΔE). We assign a “half-cell” potential to each “half-reaction”, and then add up two half-reactions to get the full reaction and full potential. We choose as a standard the reaction of hydrogen : 2H + (aq) + 2e –  H 2 (g) E 0 = 0.00 V (1 atm H 2, [H + ] = 1 M, at all temperatures) Standard Reduction Potential Indicates standard conditions

Lecture 25-6 Determining E 0 cell We measure half-cell potentials relative to this reference hydrogen (H + /H 2 ) electrode, which has E 0 ref = 0.00 V. We build voltaic cells that have this reference half-cell and another half-cell whose potential is unknown, E 0 unknown. We define: E 0 cell = E 0 cathode – E 0 anode When H 2 is oxidised, the reference electrode is the anode, so: E 0 cell = E 0 cathode – E 0 ref = E 0 unknown – 0.00 = E 0 unknown When H + is reduced, the reference electrode is the cathode, so: E 0 cell = E 0 ref – E 0 anode = E 0 unknown = -E 0 unknown

Lecture 25-7 Standard Hydrogen Electrode Experiment 1:Zn(s) + 2H + (aq)  Zn 2+ (aq) + H 2 (g) 0.76 V Potential(V) all reagents at standard concentration of 1.0 M Demo: Potential of cell Zn/H 2 Zn(s)|Zn 2+ ||H + (aq)|H 2 (g)|Pt We measure E 0 cell = 0.76 V; Since E 0 cell = 0 - E 0 unknown Then E 0 Zn = V

Lecture 25-8 Cu 2+ (s) + H 2 (g)  2H + (aq) + Cu(s) Experiment 2: V Potential(V) Demo: Potential of cell Cu/H 2 H 2 (g)| H + (aq)|| Cu 2+ (s)|Cu(s)| Pt all reagents at standard concentration of 1.0 M We measure E 0 cell =0.34 V Since E 0 cell = E 0 unknown Then E 0 Cu = 0.34 V

Lecture 25-9 E 0 Cu 0.34 V E 0 Zn V 1.10 V Potential(V) Experiment 3:Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Demo: Potential of Zn/Cu cell We measure a potential E 0 cell = 1.1 V. Zn(s)|Zn 2+ (aq)||Cu 2+ (aq)|Cu(s) E 0 cell = E 0 cathode – E 0 anode = 0.34 – (–0.76)= = 1.10 V

Lecture Cell Potential all reagents at standard concentration of 1.0 M - Experiment 1: Zn(s) + 2H + (aq)  Zn 2+ (aq) + H 2 (g) Experiment 3: Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Cu 2+ (aq) + H 2 (g)  Cu(s) + 2H + (aq) Experiment 2: -(-0.76 V) 0.34 V = 1.10 V Potential(V) Conclusion 2: You can add cell potentials as you add chemical reactions. Conclusion 2: You can add cell potentials as you add chemical reactions. Conclusion 1: Reverse the reaction – reverse the sign of the potential.

Lecture We measured for the reaction Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) A cell potential E cell = ΔE = 1.1 V Measured in volts, V = J·C –1 Moving one Coulomb of charge from Zn to Cu 2+ releases 1.1 J of energy Electromotive Force ΔE

Lecture We define the standard reduction potential E 0 as the potential of each redox couple when all reagents are in the standard state, i.e. 1 M concentration, 298 K, and gases at 1 atm pressure. The standard reduction potentials are tabulated. In data tables all half reactions are written as reductions. Fe 3+ + e –  Fe 2+ E 0 = 0.77 V Cu e –  Cu E 0 = 0.33 V 2H + + 2e –  H 2 (g) E 0 = 0.00 V Zn e –  Zn E 0 = –0.76 V The higher E (more positive), the greater the tendency to acquire electrons (be reduced). Standard Reduction Potentials

Lecture Reduction potential table Half-reactionHalf-cell potential (V) Au 3+ (aq) + 3e -  Au(s) Cl 2 (g) + 2e -  2Cl - (aq) O 2 (g) + 4H + (aq) + 4e -  2H 2 O(l) Ag + (aq) + e -  Ag(s) Cu 2+ (aq) + 2e -  Cu(s) H + (aq) + 2e -  H 2 (g) 0.00* Sn 2+ (aq) + 2e -  Sn(s) Fe 2+ (aq) + 2e -  Fe(s) Zn 2+ (aq) + 2e -  Zn(s) H 2 O(l) + 2e -  H 2 (g) + 2OH - (aq) Mg 2+ (aq) + 2e -  Mg(s) * by definition Strong oxidising agent Weak oxidising agent Strong reducing agent Weak reducing agent No/slow oxidation by H + due to an over- potential

Lecture Using the redox potential tables 1. Write down the two half-reactions. 2. Work out which is the oxidation and which is the reduction half- reaction. 3. Balance the electrons. 4. Add up the half-reactions to get full reaction. 5. Add up half-cell potentials to get E A spontaneous (working) voltaic cell, ALWAYS has a positive cell potential, E 0.

Lecture Approach Calculate the standard cell potential for a galvanic cell formed by Mg 2+ (aq) | Mg(s) in one half-cell and Sn 2+ (aq) | Sn(s) in the other. Which reaction is the oxidation and which is the reduction? Which half reaction is turned around? Sn e -  SnE 0 = -0.14V Mg e -  MgE 0 = -2.37V In general, you reverse the reaction that appears lower in the table of standard reduction potentials. In general, you reverse the reaction that appears lower in the table of standard reduction potentials.

Lecture Example 1 Calculate the standard cell potential for a galvanic cell formed by Mg 2+ (aq) | Mg(s) in one half-cell and Sn 2+ (aq) | Sn(s) in the other. We saw that Mg is a stronger reducing agent than Sn, i.e. it likes to be oxidised. Turn the Mg reaction around to get an oxidation reaction. Sn e -  SnE 0 = -0.14V Mg e -  MgE 0 = -2.37V Sn e -  SnE 0 = -0.14V Mg  Mg e - E 0 = +2.37V Mg(s) + Sn 2+ (aq)  Mg 2+ (aq) + Sn(s)E 0 = V

Lecture Example 2 Calculate the standard cell potential for a galvanic cell with Ag + (aq) | Ag(s) in one half-cell and Cr 3+ (aq) | Cr(s) in the other. E 0 (Ag+(aq)|Ag(s)) = V; E 0 (Cr 3 +(aq)|Cr(s)) = -0.74V Cr half reaction is lower (more negative), turn it around… Cr(s) + 3Ag + (aq)  Cr 3+ (aq) + 3Ag(s)E 0 = +1.54V Balance the electrons… Note! Note: E 0 does not depend on stoichiometry! Ag + + e -  AgE 0 = +0.80V Cr  Cr e - E 0 = +0.74V 3Ag + + 3e -  3AgE 0 = +0.80V Cr  Cr e - E 0 = +0.74V

Lecture Al + Fe 2+  Al 3+ + Fe ONs: Al  Al 3+ Fe 2+  Fe Oxidation half cell: Al  Al e - Reduction half cell: 2e - + Fe 2+  Fe Combine: 2 x (Al  Al e - ) 3 x (2e - + Fe 2+  Fe) 2Al + 3Fe 2+  2Al Fe Balancing Redox Equations – Example 1 (oxidation) (reduction) balance charge with e -

Lecture Cr 2 O I -  Cr 3+ + I 2 (in acidic solution) O.N.s: Cr 2 O 7 2- (+6)  Cr 3+ (+3) I - (-1)  I 2 (0) Oxidation half cell: 2I -  I 2 + 2e - Reduction half cell: 6e H + + Cr 2 O 7 2-  2Cr H 2 O Combine: 3 x (2I -  I 2 + 2e - ) 6e H + + Cr 2 O 7 2-  2Cr H 2 O 6 I H + + Cr 2 O 7 2-  3I 2 + 2Cr H 2 O (oxidation) (reduction) Balancing Redox Equations – Example 2

Lecture HgO + Zn  Hg + Zn(OH) 2 (in basic solution)  ONs: HgO (+2)  Hg (0) Zn (0)  Zn 2+ (+2)  Oxidation half cell: Zn  Zn e -  Reduction half cell: 2e - + H 2 O + HgO  Hg + 2OH - (Add OH - and water to neutralise the charge and balance O and H)  Combine: Zn  Zn e - 2e - + H 2 O + HgO  Hg + 2OH - Zn + H 2 O + HgO  Zn 2+ + Hg + 2OH - Balancing Redox Equations – Example 3 (oxidation) (reduction)

Lecture Step 1Assign O.N. to all elements. Step 2 Identify the species being reduced/oxidised. Step 3 Calculate and balance the gain/loss electrons. Step 4 Balance the number of all the atoms other than O and H. Step 5 Balance O by adding H 2 O to either side as required. Balance H and charges by adding H + to either side as required. Step 6 If basic conditions are specified, add OH – as required. (H + and OH - cannot be present at the same time, they will convert into H 2 O). Finally, check that the whole reaction is balanced, i.e that the charges and the moles of reactants and products are balanced. Balancing Redox Equations – oxidation number method

Lecture Example 1 Balance the equation for the oxidation of HCl by MnO 4 – MnO 4 – + HCl  Mn 2+ + Cl 2 Answer: 2 MnO 4 – (aq) + 10 HCl(aq) + 6 H + (aq)  2 Mn 2+ (aq) + 5 Cl 2 (aq) + 8 H 2 O(l)

Lecture Example 2 Balance the reaction of oxidation of H 2 SO 3 by Cr 2 O 7 2- : Cr 2 O H 2 SO 3  Cr 3+ + SO 4 2- Answer: Cr 2 O H 2 SO H + → 2 Cr SO H 2 O

Lecture Example Example 3 Balance the reaction between NaMnO 4 and Na 2 C 2 O 4 : MnO C 2 O 4 2-  MnO 2 + CO 3 2- Answer: 2 MnO C 2 O H 2 O → 2 MnO 2 +6 CO H +

Lecture Summary CONCEPTS  Galvanic/Voltaic cells  Cell notation  Standard Reduction Potentials  Balancing Redox Equations CALCULATIONS  Work out cell potential from reduction potentials

Lecture Half-cell standard reduction potentials