Electrochemistry

-the study of the interchange of chemical and electrical energy

-involves 2 types of redox reactions

  • generation of electrical current from a spontaneous chemical reaction
  • use of current to produce a chemical reaction (opposite of 1st one)

REVIEW OF REDOX PROBLEMS

OIL-RIGOxidation Is the Loss of electrons

Reductions Is the Gain of electrons

LEO the lion says GERLose Electrons in Oxidation

Gain Electrons in Reduction

Galvanic Cells (20.3)

Harness the energy of this reaction...

Galvanic cella device in which chemical energy is changed to electrical energy

spontaneous redox rxn producing current can be used to do work

Salt bridge/porous disktube connecting two solutions

needed to keep each compartment with a net charge of zero by

migration of ions not electrons

Anodecompartment where oxidation occurs (both begin with vowels)

AnOx--oxidation at the anode

Cathodecompartment where reduction occurs

RedCat--reduction at the cathode

Currentflow of positive charge (opposite to the flow of electrons)

Cell potential (E) /Electromotive force (emf)the driving force or pull of the electrons

--measured in volt (1 V = 1 Joule/Coulomb of charge)

Voltmeterdraws current through a known resistance – some waste from

frictional heating  measurement less than the maximum

cell potential -- digital voltmeters now used with

negligible current loss

Potentiometerdraws current without waste – variable

works against cell potential – opposite of voltmeter

Example of a Galvanic Cell:

Standard Reduction Potentials

-all half reactions are given as reduction processes in standard tables – at 1 atm for gasses & 1M for solutes – the degree sign indicates at standard state

-the standard hydrogen potential is the reference potential against which all half reactions potentials are assigned

-when a half reaction is reversed, the sign of E is reversed

-when a half reaction is multiplied by an integer, the E remains the same

-a galvanic cell runs spontaneously in the direction that gives a positive value for E

-the larger the E, the more spontaneous is the reaction

-sometimes there are only ions involved in the redox reaction; in these cases, an inert conductor, such as Pt must be used

Sample Problem1- Follow example 20.5 on pg 766.

a. Consider a galvanic cell based on the reaction

Cu2+ + Zn  Zn2+ + Cu

The half-reactions are

Cu2+ + 2e-  Cu(s)E = +0.34 V

Zn2+ + 2e- Zn(s)E = - 0.76 V

Give the balanced cell reaction and calculate E for the cell. Label the galvanic cell above with the following: Anode, cathode, direction of electron flow.

b. A galvanic cell is based on the reaction MnO4- (aq) + H+(aq) + ClO3-(aq)  ClO4-(aq) + Mn2+(aq) + H2O(l).

The half-reactions are

MnO4- + 5 e’ + 8 H+  Mn2+ + 4 H2OE = 1.516V

ClO4- + 2 H+ + 2 e’  ClO3- + H2OE = 1.19V

Give the balanced cell reaction and calculate the E for the cell.

Line Notation

-anode components on the left, cathode components on right

-separated by vertical parallel lines (indicating salt bridge or porous disk)

-phase boundary represented by a single vertical line

-anode material on far left, cathode material on far right

-catalysts are at the far ends separated by single vertical lines if only dissolved materials are present

Example: Pt(s) ClO3-(aq) , ClO4-(aq) MnO4-(aq) , Mn2+(aq) Pt(s)

Sample Problem 2 – Describe completely the galvanic cell based on the following half reactions under standard conditions:

Ag+ + e- AgE = 0.80 V(1)

Fe3+ + e- Fe2+E = 0.77 V(2)

Cell Potential, Electrical Work, and Free Energy

potential difference = emf (electromotive force) = voltage (V) = work (J) / charge (C)

  • work depends on the push (thermodynamic driving force) behind the electrons
  • the driving force is the emf: potential difference between 2 points

w = -q E

w = work (J)

E =cell potential, emf

q = nF

q = quantity of charge transferred (C)

n = moles of electrons

F = Faraday’s constant = 96,485 C/mol of electrons

At standard conditions, Go = -nF E o

  • the maximum cell potential is directly proportional to the change in free energy between the reactants and products in the cell
  • this equation confirms that a galvanic cell runs in the direction that gives a positive cell potential
  • the E o of the cell correlates to a negative G, which means the reaction is spontaneous

In real spontaneous processes, some energy is wasted; work never is the maximum possible if any current is flowing; lost to frictianl heating

efficiency = w / wmax x 100

Sampkle Problem 17.3: Calculate Go for the following reaction: Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq)

Is the rxn spontaneous?

Smapel Problem 17.4: Preidct whether 1 M HNO3 will dissolve gold metal to form 1 M Au3+ solution.

The Dependence of Cell Potential on Concentration

If the cell concentration is not equal, use Le Chatelier’s principle to predict the change that will take place.

Ex: Cu(s) + 2 Ce4+(aq)  Cu2+(aq) + 2 Ce3+(aq)

  • as the concentration of Ce4+ increases, the reaction will proceed to the products’ side and increase the driving force on the electrons

Sample Problem 17.5: For the cell reaction 2 Al(s) +3 Mn2+(aq)  2 Al3+(aq) + 3 Mn(aq) E o cell= 0.48 V

Predict whether the E o cell is larger or smaller than in the following cases.

  1. [Al3+] = 2.0 M, [Mn2+] = 1.0M
  1. [Al3+] = 1.0 M, [Mn2+] = 3.0M
Concentration Cells
  • Galvanic cells can be made where the concentration in each compartment is different but with the same ions
  • Nature tries to make both sides equal, so electrons will go to the side with the higher concentration
  • In these types of cells, the voltage is small – concentration is the only factor producing the cell potential

Ex: Figure 17.9

Sample Problem 17.6: Determine the direction of electron flow and designate the anode and cathode for the cell represented in Figure 17.10.

The Nerst Equation

Electrochemistry (Chapter 20)1