8 - 1 Electrochemistry Electrochemistry deals with the conversions between electrical and chemical energy. In particular, the chemical reactions are oxidation-reduction reactions.  Oxidation reactions give off electrons and what is being oxidized is called the reducing agent.  Reduction reacts take in electrons and what is being reduced is called the oxidizing agent.

8 - 2  Remember, LEO goes GER. LEO is the loss of electrons - oxidation. GER is the gain of electrons - reduction. An electrochemical cell (also called a voltaic or galvanic cell) converts chemical energy to electrical energy.

8 - 3 An Electrochemical Cell V voltmeter salt bridge Zn (anode/-) Cu (cathode/+) electrolyte (NaNO 3 ) (Na +, NO 3 - ) ZnSO 4 (Zn 2+, SO 4 2- )CuSO 4 (Cu 2+, SO 4 2- )

8 - 4 The following is what should be known about this particular voltaic cell:  The oxidation occurs in the anode compartment. The anode reaction is Zn → Zn e -. Because electrons are not good swimmers, they collect on the metal strip which is called the anode (a negative electrode).

8 - 5  The reduction occurs in the cathode compartment. The cathode reaction is Cu e - → Cu. The cathode is assigned a positive charge.  A voltmeter connected to the two electrodes will measure the electromotive force (emf) of the cell.

8 - 6 Electromotive force (E) is really a misnomer because what is being measured is voltage. Voltage really measures the electric potential energy in joules/coulomb, 1 V = J/C.  Electrons will flow from negative to positive (anode to cathode) in the external circuit.

8 - 7  Internally within the cell itself, the cations (positive ions) move to the cathode and the anions move to the anode.  The salt bridge allows for ions to flow in and out of each cell compartment. A salt is chosen so that it’s ions, in this case Na + and NO 3 -, will not react with the species in either compartment. The salt bridge allows for electric neutrality.

8 - 8  The Na + provided by the salt bridge migrate to the cathode to compensate for the reduction of cations. The NO 3 - migrate to the anode to compensate for the increasing concentration of cations.  Standard state conditions for a voltaic cell exist when gases are at a pressure of 1 atm, all solutes are at a concentration of 1 M, and T = 25°C = 298K.

8 - 9  All aqueous solutions must have a concentration of 1.0 M, a pressure of 1.0 atm and a temperature of 25°C.  As previously stated, the term force is somewhat of a misnomer. Because the anode is considered at a higher potential than the cathode, what actually is being measured is electrical potential energy in volts, 1 V = 1 J 1 C

What Is The Voltmeter Measuring? A voltmeter measures the cell emf which measures the potential difference between the anode and cathode.  The reference electrode which all other electrodes are compared to is the SHE (Standard Hydrogen Electrode).  2H + (aq) + 2e - → H 2 (g)E° = 0.00 V

 The cell potential for any combination of electrodes is given by E° cell = E° ox + E° red

Predicting Spontaneous Reactions (a)Is the following reaction spontaneous? If it is spontaneous, determine the cell emf and identify the anode and cathode. Zn + 2H + → Zn 2+ + H 2 Zn → Zn e - E° ox = 0.76 V 2H + + 2e - → H 2 E° red = 0.00 V Zn(s) + 2H + (aq) → Zn 2+ (aq) + H 2 (g) E° = 0.76 V

The reaction is spontaneous because E° > 0. Zn is being oxidized making Zn the reducing agent or reductant. H + is being reduced making H + the oxidizing agent or oxidant. It did not occur in this problem but if you had to multiply the number of electrons by a coefficient to balance them, you would not multiply the E° by the coefficient.

The mass of the electrodes change as the oxidation-reduction reaction proceeds.  At the anode, zinc atoms are consumed as they are oxidized into Zn 2+ cations which decreases the mass of the Zn anode.  At the cathode H + cations are reduced to H 2 molecules which increases the mass of the accumulated hydrogen.

(b) Is the following reaction spontaneous? If it is spontaneous, determine the cell emf and identify the anode and cathode. Li + + Na → Na + + Li Na → Na + + e - E° = 2.71 V Li + + e - → LiE° = V E° = V

The reaction is not spontaneous because E° < 0. The reverse direction is spontaneous and is given by Na + (aq) + Li(s) → Na(s) + Li + (aq) E° = 0.33 V Li is being oxidized making Li the reducing agent or reductant. Na + is being reduced making Na + the oxidizing agent or oxidant.

Sometimes a line convention is used to identify the anode, salt bridge, and cathode as shown below.  Zn|Zn 2+ (aq)||H + (aq)|H 2 The anode components are written to the left of the salt bridge symbol (||) and cathode components are written to the right. The single vertical line indicates a phase boundary as in the case of (s) and (aq).

Determine which of the following metals Ni, Cr, and Ca is the most active metal. Ca 2+ (aq) + 2e - → Ca(s)E° red = V Cr 3+ (aq) + 3e - → Cr(s)E° red = V Ni 2+ (aq) + 2e - → Ca(s)E° red = V The more active the metal, the more it wants to give off electrons. Reversing the reduction reactions gives E° ox = 2.87 V (TP!) corresponding to calcium making it the most active metal.

SRP Summary Think Positive! The more positive the standard reduction potential, E° red, the more the species wants to be reduced making it a stronger oxidizing agent.  The stronger the oxidizing agent the weaker the conjugate reducing agent.  F 2 is the strongest oxidizing agent because it is the most easily reduced, E° red = 2.87 V.

If F 2 is the easiest species to reduce, then F - must be the hardest to oxidize. Although this logic may sound twisted, it is very similar to that of a strong acid-weak conjugate base pair.  The stronger the reducing agent the weaker the conjugate oxidizing agent. The more negative the standard reduction potential, E° red, the more the species wants to be oxidized making it a stronger reducing agent.

 The stronger the reducing agent the weaker the conjugate oxidizing agent.  Li is the strongest reducing agent because it is the most easily oxidized, E° ox = 3.05 V.  If Li is the easiest species to oxidize, then Li + must be the hardest to reduce.

Spontaneity, E°, and Δ G° When there is a spontaneous redox reaction, E° > 0. The equation relating Δ G° (Gibb’s free energy) and cell potential is given by Δ G° = -nFE° where n is the number of moles of electrons transferred and F is Faraday’s constant

and Remember that the superscript ° indicates if all the products are in their standard states, the P = 1 atm, the concentrations are 1.0 M and T = 25°C = 298K. Many times redox reactions will not be run at standard state conditions and you will have to use the Nernst equation given by F = J V mol e -.

E = E° V n log Q where E° is the cell potential at standard conditions, Q is the reaction quotient, n is the number of moles of electrons, and T = 25°C = 298K. Remember, the concentrations of solids, liquids, and solvent do not appear in the reaction quotient.

If a redox reaction reaches equilibrium, Δ G = 0 and E = 0 and at 298K the following is true E° = V n log K The emf is not converted to work with 100% efficiency. There is always energy wasted in the form of heat with Δ G° being the work attainable.

Calculate Δ G in an Alkaline Solution Calculate the standard free energy, Δ G, for the following redox reaction in a basic solution. Ni 2+ (aq) + Cr(OH) 3 (s) → Ni(s) + CrO 4 2- (aq) Ni e - → Ni Cr(OH) 3 → CrO

Cr(OH) 3 → CrO e - Cr(OH) 3 + H 2 O → CrO e - + 5H + 3Ni e - → 3Ni 2Cr(OH) 3 + 2H 2 O → 2CrO e H + 3Ni 2+ +2Cr(OH) 3 + 2H 2 O → 3Ni + 2CrO H + 10OH - 3Ni 2+ (aq) +2Cr(OH) 3 (s) + 10OH - (aq) → 3Ni(s) + 2CrO 4 2- (aq) + 8H 2 O(l) E° cell = E° red + E° ox = V V = V

Δ G° = -nFE° Δ G° = - 6 mol e - × Δ G° = 8.7 x 10 4 J The reaction is not spontaneous in the forward direction but is so is the reverse direction C 1 mol e - × V

The Nernst Equation Revisited As an electrochemical cell operates, it is always trying to achieve chemical equilibrium. When the cell potential is measured at the instant the spontaneous redox reaction begins, two conditions exist:  The cell potential is a maximum.  The system is the furthest from equilibrium as it gets.

For a given set of electrodes at 25°C = 298K, cell potential is dependent on concentration.  An increase in reactant and a decrease in product concentrations increases the cell potential. At 25°C = 298K, the Nernst equation (which was presented on slides 23-25) calculates the voltage when the concentrations are not 1.0 M.

Using The Nernst Equation For the following reaction taking place in a voltaic cell at standard temperature, 2Al(s) + 3I 2 (s) → 2Al 3+ (aq) + 6I - (aq) Determine the emf when [Al 3+ ] = 4.0 x M and [I - ] = M.

Al → Al e - 3I 2 + 6e - → 6I - 2Al → 2Al e - E° ox = 1.66 V 3I 2 + 6e - → 6I - E° red = 0.54 V 2Al(s) + 3I 2 (s) → 2Al 3+ (aq) + 6I - (aq) E° = 2.20 V E = E° V n log Q

E = 2.20 V V 6 log ((4.0 x ) 2 × ) - × = 2.37 V As the cell continues to spontaneously discharge, the concentrations change and E° of the cell decreases. Voltaic cells spontaneously discharge until equilibrium is achieved, at which point the cell is dead.

Electrolytic (Daniel) Cell A nonspontaneous redox reaction can be forced to occur by connecting the electrodes to a source of direct current (DC) electricity. An electrolytic cell converts electrical energy to chemical energy. This process is called electrolysis. An electrolytic cell is diagrammed on the next slide.

An Electrolytic Cell direct current power source salt bridge Zn (cathode/-) Cu (anode/+) electrolyte (NaNO 3 ) (Na +, NO 3 - ) ZnSO 4 (Zn 2+, SO 4 2- )CuSO 4 (Cu 2+, SO 4 2- ) +-

Components of an Electrolytic Cell At first glance, it is easy to confuse an electrolytic cell with an electrochemical cell. For an electrolytic cell:  A source of direct current replaces the voltmeter or electrical appliance.  Electrons are forced onto the cathode making the cathode negative.

 Electrons are removed from the anode making it positive.  The cathode being the negative electrode and the anode being the positive anode is the opposite of the electrode charge found in a voltaic cell.  The cathode is still the site of the reduction and the anode the site of the oxidation.

The reduction reaction is given by Zn 2+ (aq) + 2e - → Zn(s) and the anode reaction is given by Cu(s) → Cu 2+ (aq) + 2e -  In an electrochemical cell using zinc and copper electrodes, the E° cell = 1.10 V. This requires of the direct current source to be greater than 1.10 V.

Electrolytical Cell Terminology. I = Q t I is the current measured in amps (A), Q is the magnitude of charge in coulombs (C), and t is the time in seconds (s). 1 A = 1C 1s

The charge on 1 e - = 1.6 x C 1 mol e - = 6.02 x e - ≈ C The quantity of charge can also be expressed in faradays, F, where 1 F = C = 1 mol e -. Whenever work is done on any system, there is an increase of energy, whether it be kinetic or potential.  In the case of an electrolytic cell, there will be an increase in electric potential energy.

P = W t = ΔEΔE t = J s = watts 1 J = 1 watt-s or 3.6 x 10 6 J = 1 kWh

Electrolytic Problems An acidic solution contains Pb 2+ cations and an anion which is a spectator. When the solution is electrolyzed, PbO 2 is formed at the anode and Pb at the cathode. (a)What reaction occurs at the anode? The oxidation of Pb 2+ occurs at the anode and the reaction is Pb 2+ (aq) + 2H 2 O(l) → PbO 2 (s) + 4H + (aq) + 2e -

(b) If a current of 1.25 A is applied to the cell for 15.0 min, what mass of Pb is deposited at the cathode? I = 1.25 A t = 15.0 min 1 mol e - = C m = 1.25 C s × 15.0 min × 60 s 1 min × 1 mol e C × 1 mol Pb 4+ 2 mol e - × g Pb 4+ 1 mol Pb 4+ = 1.21 g Pb 4+