Standard Reference Electrode Standard Hydrogen Electrode (SHE) SHE: • Assigned 0.000 V • Can be anode or cathode • Pt does not take part in reaction • Difficult to operate Standard Conditions: 1 atm for gases, 1.0M for solutions, 25 oC for all (298 K)
Alternative reference electrodes • Ag/AgCl electrode AgCl(s) + e- Cl- + Ag(s) Ecell = +0.22 V vs. SHE • Calomel electrode Hg2Cl2(s) + 2e- 2Cl- + 2Hg(l) Ecell = + 0.24 V vs.SHE
Standard Reduction Potentials in Aqueous Solution at 25° C How can we determine which substance is being oxidized and which is being reduced? …..The MORE POSITIVE reduction potential gets to be reduced.
Zn CuSO4 1 hour Reading the reduction potential chart • Elements that have the most positive reduction potentials are easily reduced. • Elements that have the least positive reduction potentials are easily oxidized. • The table can also be used to tell the strength of various oxidizing and reducing agents. • It can also be used as an activity series. Metals having less positive reduction potentials are more active and will replace metals with more positive potentials.
e- V e- Zn Cu Zn2+ Cu2+ NO3- NO3- Calculating Standard Cell Potential IMPORTANT: Write both equations AS IS from the table in the reduction form with their voltages. Example: Calculate E0 for the cell shown in the Figure below: Cu2+ + 2e- Cu(s)Eo = 0.337 V Zn2+ + 2e- Zn(s)Eo = -0.763 V Eo = (0.337) – (-0.763) = +1.10 V
Remember: • Cu should be the cathode (it has higher Eo). • Oxidation occurs at the anode (may show mass decrease). • Reduction occurs at the cathode (may show mass increase). • In terms of electrode charge • Electrolytic cell Anode (+) Cathode (-) • Galvanic cell Anode (-) Cathode (+) • (°) means standard conditions: 1atm, 1M, 25C. • Negative Eo implies non-spontaneous • Positive Eo implies spontaneous (would be a good battery!).
Example: Will the following mixture react spontaneously at standard state? Zn2+(aq) + Fe2+(aq) ??? Given: • Zn2+ + 2e- Zn Eo= -0.76 V • (b) Fe3+ + e- Fe2+ Eo = 0.77 V • -2(b): Zn2+ + Fe2+ Zn(s) + 2Fe3+ E0 for the overall reaction = -0.76 – (0.77) = -1.53 V The forward reaction is not spontaneous. Example: Consider a galvanic cell based on the reaction: Ag+(aq) + Sn(s) → Ag(s) + Sn2+(aq) Give the balanced cell reaction and calculate E° for the cell. Ag+ + e- Ag(s)Eo = 0.80 V Sn2+ + 2e- Sn(s)Eo = -0.14 V 2Ag+ + Sn 2Ag(s) + Sn2+ Eo = 0.80 – (-0.14) = 0.94 V
Example: Using the table of standard reduction potentials, predict whether 1 M HNO3 will dissolve gold metal to form a 1 M Au3+ solution. …… ….. No! Au3+ + 3e- Au(s) Eo = 1.50 V NO3- +4H3O+ + 3e- NO(g) + 6H2O(l) Eo = 0.96 V To dissolve Au, it should be as Au3+ NO3- +4H3O+ + Au Au3++ NO(g) + 6H2O(l) Eo = ?? V
Dependence of Cell Potential on Concentration Voltaic cells at non-standard conditions -- LeChatlier’s principle can be applied. An increase in the concentration of a reactant will favor the forward reaction and the cell potential will increase. The converse is also true! Example: For the cell reaction: 2Al + 3Mn2+→ 2Al3+ + 3Mn E°cell = ?? Predict whether Ecell is larger or smaller than E°cell for the following cases: a. [Al3+ ] = 2.0 M, [Mn2+ ] = 1.0 M b. [Al3+ ] = 1.0 M, [Mn2+] = 3.0 M A: Ecell < E°cell B: Ecell > E°cell
When cell is not at standard conditions, use Nernst Equation aA + bB cC + dD • In a chemical reaction such as: Substitute Where: Go : Free energy change when all the reactants and products are in their standard states (unit activity). R : is the gas constant. T : is the temperature in the absolute temperature Q : Reaction Quotient
K …… Nernst Equation • Where concentrations are substituted for activities • At 298 K the equation becomes At Equilibrium, G = 0, E = 0. Hence
Concentration Cells We can construct a cell where both compartments contain the same components BUT at different concentrations. In the picture,Silver will be deposited on the right electrode, thus lowering the concentration of Ag+ in the right compartment. In the left compartment the silver electrode dissolves [producing Ag+ ions] to raise the concentration of Ag+ in solution. Example: Using the table of standard reduction potentials, calculate ∆G° for the reaction: Cu2+ + Fe → Cu + Fe2+ • Is this reaction spontaneous? ………. Yes!
Example : Determine Eocell and Ecell based on the following half-reactions: VO2+ + 2H+ + e-→ VO2+ + H2O E° = 1.00 VZn2+ + 2e- → Zn E° = -0.76VWhere: 25°C, [VO2+] = 2.0 M, [H+] = 0.50 M, [VO2+] = 1.0 x 10-2M, [Zn2+] = 1.0 x 10-1M = 1.00 – (-0.76) = 1.76 V 2VO2+ + 4H+ + Zn → 2VO2+ + 2H2O + Zn2+
Example: Calculate Kw, the ion-product constant of water, using the given data. H2O + e- 1/2H2 + OH- Eo = -0.83 V H+ + e- 1/2H2 Eo = 0.00 V Should be Anode Therefore, we need this equation: the ion-product H2O H+ + OH- At equilibrium, Ecell = 0, and [H+][OH-] = Kw Kw = 1 x 10-14