# 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.

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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 o C for all (298 K)

Alternative reference electrodes Ag/AgCl electrode AgCl (s) + e -  Cl - + Ag (s) E cell = +0.22 V vs. SHE Calomel electrode Hg 2 Cl 2(s) + 2e -  2Cl - + 2Hg (l) E cell = + 0.24 V vs.SHE

How can we determine which substance is being oxidized and which is being reduced? …..The MORE POSITIVE reduction potential gets to be reduced. Standard Reduction Potentials in Aqueous Solution at 25° C

 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. Reading the reduction potential chart 1 hour Zn CuSO4

Calculating Standard Cell Potential IMPORTANT: Write both equations AS IS from the table in the reduction form with their voltages. Cu 2+ V NO 3 - Zn 2 + NO 3 - Cu Zn e-e- e-e- Example: Calculate E 0 for the cell shown in the Figure below: Zn 2+ + 2e -  Zn (s) E o = -0.763 V Cu 2+ + 2e -  Cu (s) E o = 0.337 V E o = (0.337) – (-0.763) = +1.10 V

Remember: -Cu should be the cathode (it has higher E o ). - 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 cellAnode (-)Cathode (+) - (°) means standard conditions: 1atm, 1M, 25  C. - Negative E o implies non-spontaneous - Positive E o implies spontaneous (would be a good battery!).

Example: Consider a galvanic cell based on the reaction: Ag + (aq) + Sn(s) → Ag(s) + Sn 2+ (aq) Give the balanced cell reaction and calculate E° for the cell. E o = 0.80 – (-0.14) = 0.94 V Sn 2+ + 2e -  Sn (s) E o = -0.14 V Ag + + e -  Ag (s) E o = 0.80 V 2Ag + + Sn  2Ag (s) + Sn 2+ Example: Will the following mixture react spontaneously at standard state? Zn 2+ (aq) + Fe 2+ (aq)  ??? Given: (a)Zn 2+ + 2e -  ZnE o = -0.76 V (b) Fe 3+ + e -  Fe 2+ E o = 0.77 V (a)-2(b): Zn 2+ + Fe 2+  Zn (s) + 2Fe 3+ E 0 for the overall reaction = -0.76 – (0.77) = -1.53 V The forward reaction is not spontaneous.

Example : Using the table of standard reduction potentials, predict whether 1 M HNO3 will dissolve gold metal to form a 1 M Au 3+ solution. …… ….. No! Au 3+ + 3e -  Au (s) E o = 1.50 V NO 3 - +4H 3 O + + 3e -  NO (g) + 6H 2 O (l) E o = 0.96 V To dissolve Au, it should be as Au 3+ NO 3 - +4H 3 O + + Au  Au 3+ + NO (g) + 6H 2 O (l) E o = ?? 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 + 3Mn 2+ → 2Al 3+ + 3Mn E° cell = ?? Predict whether E cell is larger or smaller than E° cell for the following cases: a. [Al 3+ ] = 2.0 M, [Mn 2+ ] = 1.0 M b. [Al 3+ ] = 1.0 M, [Mn 2+ ] = 3.0 M A: E cell < E° cell B: E cell > E° cell

In a chemical reaction such as: aA + bB  cC + dD Where:  G o : 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 When cell is not at standard conditions, use Nernst Equation Substitute

 Where concentrations are substituted for activities  At 298 K the equation becomes  At Equilibrium,  G = 0, E = 0. Hence K …… Nernst Equation

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: Cu 2+ + Fe → Cu + Fe 2+ Is this reaction spontaneous? ………. Yes!

VO 2 + + 2H + + e - → VO 2+ + H 2 O E° = 1.00 V Zn 2+ + 2e - → Zn E° = -0.76V Where: 25°C, [VO 2 + ] = 2.0 M, [H + ] = 0.50 M, [VO 2+ ] = 1.0 x 10 -2 M, [Zn 2+ ] = 1.0 x 10 -1 M Example : Determine E o cell and E cell based on the following half- reactions: = 1.00 – (-0.76) = 1.76 V 2VO 2 + + 4H + + Zn → 2VO 2+ + 2H 2 O + Zn 2+

Example: Calculate Kw, the ion-product constant of water, using the given data. H 2 O + e -  1/2H 2 + OH - E o = -0.83 V H + + e -  1/2H 2 E o = 0.00 V H 2 O  H + + OH - Therefore, we need this equation: the ion-product Should be Anode At equilibrium, E cell = 0, and [H + ][OH - ] = K w K w = 1 x 10 -14

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