electrochemistry
TRANSCRIPT
Electrochemistry Physical Chemistry
Daniel Cell1. Electrochemistry is the study of the interconversion of electrical and
chemistry energy. 2. Voltaic cell, in which a spontaneous reaction generates electrical energy.3. Electrolytic cell: electrical energy is used to bring about a non spontaneous
reaction.
Oxidation-reduction (redox) reactions occur when electrons are given up by the substance being oxidized (the reducing agent) and simultaneously gained by the substance being reduced (the oxidizing agent).
2 Ag+ + Zn Zn2+ + 2 Ag
At Anode : Zn Zn2+ + 2e
At Cathode: Ag+ + e Ag
1. The two half-cells are connected by a salt bridge, which prevents free mixing of both electrolyte solutions but permits the proper movement of ions to maintain electrical neutrality.
2. When Zn is oxidized some anions must enter (or cations must leave) the Zn half-cell to compensate for the added positive charge of the Zn2+ produced. Also when Ag+ reduced cations must enter (or anions must leave) the Ag half-cell.
3. Let’s say the salt bridge used is: KNO3 , K+ ions move from the salt bridge into cathode half bridge. At the same time NO3
- ions move into the anode half cell.
4. The Daniel Cell Diagram: Zn(s)/ZnSO4 (aq) AgSO4(aq)/Ag(s)
Daniel Cell
1. When chlorine gas is bubbled through a water solution of NaBr, a spontaneous redox reaction occurs:
Cl2 (g) + 2Br-(aq) 2Cl-(aq) + Br2 (g)
a) What is the cathode and anode reaction?b) Which way do electrons move in the external circuit?c) Which way do anions and cations move within the cell?
Solution:a) Cathode: Cl2 + 2e 2Cl- Anode: 2Br- Br2 + 2e
b) From anode to cathodec) Anion to anode, cations to cathode
Practice
Oxidation Number Determine the oxidation number of the underlined atom:a) V2O5
b) SO42--
c) SO32-
d) OCl-
e) HCO3-
f) MnO4-
Writing Redox Equation
1. xA + yB Product x(+a) + y(-b) = 0
Practice:1. What are the oxidation numbers of KIO3, KIO, Na2SO3 and Na2SO4. Thus,
balance the equation.
xKIO3 + y Na2SO3 cKIO + d Na2SO4
2. Use the oxidation number method to balance the equation:a) Sn +HNO3 SnO2 + NO2 + H2O
Changes in Oxidation Number
Writing Half EquationClO3
- + I- Cl- + I2
1. Determine change of oxidation number: ClO3
- Cl- : +5 to -1 = -6….. Reduction I- I2 : -1 to 0 = +1…. Oxidation
2. ClO3- + 6e Cl-
2I- I2 + 2e
3. H2O, H+ and OH- can be added to balance up the half equatons.
ClO3- + 6H+ + 6e Cl- + 3H2O
2I- I2 + 2e ) x 3 ---------------------------------------------------------
ClO3- + 6H+ + 6I- Cl- + 3H2O + 3I2
1. Cr2O72- + H+ + NO2- Cr3+ + NO3
-
2. Cl2 + SO2 Cl- + SO42-
3. C + H2SO4 CO2 + SO2
4. H2O2 + Mn2+ MnO2 + H2O5. Cu + HNO3 Cu2+ + NO
Practice 2
The Electrode Potential of Metal1. Suppose you have a piece of magnesium in a beaker of water.
There will be some tendency for the magnesium atoms to shed electrons and go into solution as magnesium ions. The electrons will be left behind on the magnesium
Electrode Potential
In a very short time, there will be a build-up of electrons on the magnesium, and it will be surrounded in the solution by a layer of positive ions (Helmholtz double layer). This produces a potential defference between the metal and the solution.
The Electrode Potential of Metal1. Dynamic equilibrium will be established when the rate at
which ions are leaving the surface is exactly equal to the rate at which they are joining it again. At that point there will be a constant negative charge on the magnesium, and a constant number of magnesium ions present in the solution around it.
The potential difference produced between the electrode and the solution in half cell is called electrode potential. The magnitude of the electrode potential depends on the tendency of the metal to lose electrons and to form hydrated metal ions
The Electrode Potential of Metal 1. Copper is less reactive and so forms its ions less readily. The position of the magnesium equilibrium . . .
. . . lies further to the left than that of the copper equilibrium.
The Electrode Potential of Metal All we need to know is that the magnesium equilibrium lies
further to the left than the copper one. We need to know that magnesium sheds electrons and forms ions more readily than copper does.
That means that we don't need to be able to measure the absolute voltage between the metal and the solution. It is enough to compare the voltage with a standardized system called a reference electrode.
The system used is called a standard hydrogen electrode.
The Electrode Potential of Non Metal The standard hydrogen electrode The standard hydrogen electrode looks like this:
As the hydrogen gas flows over the porous platinum, an equilibrium is set up between hydrogen molecules and hydrogen ions in solution. The reaction is catalyzed by the platinum.
• The ions in the solution have a concentrated of 1.0 M
• All gases involved have a pressure of 1 atm.
•A temperature of 25oC
Using the standard hydrogen electrode1. The standard hydrogen electrode is attached to the electrode system you are
investigating - for example, a piece of magnesium in a solution containing magnesium ions.
Using the standard hydrogen electrode1. The emf for the cell is 2.37 volt.2. According to IUPAC, the standard potential for the magnesium
electrode is -2.37. The negative sign is used to show that the electrode is the negative pole.
3. The voltmeter shows that the electrons flow from the magnesium electrode to the hydrogen electrode in the external circuit. This means that at Mg electrode:
Mg Mg2+ + 2e 3. At hydrogen electrode: 2H+ + 2e H2
4. The overall cell reaction: Mg + 2H+ Mg2+ + H2
5. The cell diagram: Mg(s)/Mg2+(aq, 1M) H+ (aq, 1M)/ H2 (g) /Pt(s)
EӨ = -2.37 V
Using the standard hydrogen electrode1. Half equation is written as a reduction process: Oxidized species + ne reduced species.2. Thus, Ag+(aq)/Ag(s) EӨ = +0.80 V means that the following half cell
reaction has a standard electrode potential of +0.80 Ag+ + e Ag EӨ = +0.80 V 3. The cell diagram is written with:a) The half cell undergoing oxidation ( negative electrode) on the left of the
diagram.b) The half cell undergoing reduction ( positive electrode) on the right of the
diagrammetal / metal ion combination E° (volts)
Mg2+ / Mg -2.37
Zn2+ / Zn -0.76
Cu2+ / Cu +0.34
Ag+ / Ag +0.80
Practice 1. Consider the following equilibrium involving I2 and Fe3+:
I2 + 2e 2I- EӨ = +0.54 V
Fe3+ + e Fe2+ EӨ = +0.77 V a) Write the ionic equation for the overall cell reaction.b) State the observation after the half cell have been connected for a short
while.c) Determine the polarity of the electrode in the electrochemical cell.Solutionsa) 2Fe3+ + 2I- 2Fe2+ + I2
b) The beaker on the right (Fe3+) : yellow to green The beaker on the left (I-) : yellow to brownc) I2 + 2e 2I- is anode cause oxidation happen.
Fe3+ + e Fe2+ is cathode cause reduction occur.
We could therefore write down the emf of the above cell (under standard conditions) as:
emf of cell = potential of Cu - potential of Zn or conventionally
For example, one could use the reaction
to construct the cell
The electrochemical Series1. The electrochemical series shows the ease of a species accept or loses
electrons.2. The higher a species the easier it loses electron (oxidized)3. The EӨ value shows the possibility of a reaction but not the quantity of
substances involved. 4. If EӨ for I2 + 2e 2I- +0.54 V, then ½ I2 + e I- not (1/2 x 0.54V) but is
+0.54V.5. The reactivity of metals increases as the electrode potential become more
negative while the reactivity if non metal increases as the electrode potential become more positive.
6. The strength of oxidizing agents increases on descending the series while the strength of reducing agents increases on ascending the series.
Practice 1. Arrange the following substances in the ascending order of their
oxidizing strength:a) Na, Cl2, Cu
b) Zn, Ag+, Agc) Cl2, Br2, Br-
2. The EӨ value for three elements, X, Y and Z are given below: X X+ + e EӨ = -0.34V Y Y+ + e EӨ = -0.80V Z Z+ + e EӨ = +2.46V
Predicting the Possibility of Redox Reaction.1. The possibility of a spontaneous reaction occurring can be predicted by
using: a) Electrochemical seriesb) EӨ value for the overall reaction
2. An oxidizing agent on the left of the electrochemical series will oxidize a reducing agent on the right if the reducing agent is above the oxidizing agent in the series.
I2 + 2e 2I- EӨ = +0.54 V
Br2 + 2e 2Br- EӨ = +1.07 V
Cl2 + 2e 2Cl- EӨ = +1.33 V
BrBr22 is able to oxidize I- into I is able to oxidize I- into I2 2 but cannot oxidize Cl- into Clbut cannot oxidize Cl- into Cl22
Predicting the Possibility of Redox Reaction.1. The possibility of a spontaneous reaction occurring can be predicted by
using: a) EӨ value for the overall reaction
2. If EӨ for a combined reaction is positive, then the reaction will occur spontaneously.
Sn2+ + 2Fe3+ Sn4+ + 2Fe2+
Sn2+ Sn4+ + 2e EӨ = -0.15 V2Fe3+ + 2e 2Fe2+ EӨ = +0.77V--------------------------------------------------------------- EӨ = +0.62V---------------------------------------------------------------
PracticePractice1. Predict whether the following reactions can take place under
standard condition.a) Reduction of Sn2+ to Sn by Mgb) Oxidation of Cl- to Cl2 acidified Cr2O7
2-
Hints:Mg2+ + 2e Mg EӨ = -2.38 V Sn2+ + 2e Sn EӨ = -0.14 V--------------------------------------------------------------------Cr2O7
2- + 14H+ + 6e 2Cr3+ + 7H2O EӨ = +1.33 V
Cl2 + 2e Cl- EӨ = +1.36 V
Predicting the Stability of Aqueous IonsPredicting the Stability of Aqueous Ions1. The relative stability of Cr and Cr2+ can be explain as follow: Cr2+ + 2e Cr EӨ = -0.91 V 2H+ + 2e H2 EӨ = 0.00 V
------------------------------------------------------- Cr + 2H+ Cr2+ + H2 EӨ = + 0.91V
-------------------------------------------------------2. This implies that Cr2+ is more stable than Cr because the forward reaction
can occur spontaneously.3. The Co3+ ion can be stabilized in aqueous solution by complexing it with
other ligands like ammonia.
The Nernst Equations 1. The value of the electrode potential depends on:a) The concentration of the ion and the temperature of the
solutionb) The pH value (acidity) of the solutionc) The effect of complex formation.For the electrode system: oxidized species + ne reduced species
E = EӨ + RT/nF {ln [oxidized species]/[reduced species]} or E = EӨ + 0.059/n {lg [oxidized species]/[reduced species]}
The Nernst Equations a A + b B = c C + d D 1. Q = [C]c[D]d/ [A]a[B]b
2. E = EӨ - 0.059/n {lg Q}
Sn2+(aq) + Br2 (l) Sn4+
(aq) + 2 Br-(aq)
Q = [Sn4+][Br-]2 / [Sn2+] The Nernst equation for this reaction is:
E = 0.092 – 0.059/2 lg[ (0.0001)(0.0001)2/ 0.05]
E = 0.92 V - (-0.316 V) E = 1.24 V
Effect of the Concentration of an ion on the emf of a cell
1. Cu(s) / Cu2+ (aq) // Ag+ (aq) / Ag (s) Cu + 2Ag+ 2Ag + Cu2+
E = EӨ - 0.059/n {lg Q} If the concentration of Cu2+ ions in the copper half cell is reduced, the ECu is
less than EӨ and the emf of the electrochemical cell increases. If the concentration of Cu2+ ions in the copper half cell is reduced, the
equilibrium will shift to the right.
The Nernst Equation and Solubility Product1. When a silver electrode is immersed into a saturated silver chloride
solution, the electrode potential at 298 K is 0.51 V. Calculate the solubility product of silver chloride at 298 K. EӨ
Ag+/Ag = +0.80 V.
E = EӨ + 0.059/1 lg[Ag+] 0.051 = 0.80 + 0.059 lg[Ag+] [Ag+] = 1.22 x 10-5 moldm-3
Ksp = [Ag+][Cl-] = (1.22 x 10-5 )2
= 1.49 x 10 -10 mol2dm-6
The Nernst Equation and Solubility Product1. Consider the cell: Pt(s) /H2 (g, 1.0 atm)/HCl(0.01M)// AgCl(aq)/Ag(s) The measured cell emf is 0.046 V at 25oC and EӨ
Ag+/Ag = +0.80 V
a) Write equation for the reaction occurring at the electrodes.b) Write an equation for the overall cell reaction.c) Calculate the emf of the cell under standard conditions.d) Calculate the [Ag+] in equilibrium with AgCl and Cl-.e) What is the concentration of Cl-?f) Find the Ksp of AgCl.
The Nernst Equation and Equilibrium Constant1. When the redox reaction in a Daniel cell has achieved equilibrium, Ecell =
0.0 V. This is because at equilibrium, there is no overall change taking place in a chemical reaction.
0 = EӨ - 0.059/n lg Kc EӨ = 0.059/n [lg Kc]
2. Consider the following reaction: 2Fe3+ + Cu 2Fe2+ + Cu2+
a) Calculate the EӨ the system.b) Calculate the equilibrium constant Kc for the reaction. [ 0.43V, 3.8 x1014]
The Nernst Equation and pH1. The cell reaction: Zn + 2H+ Zn2+ + H2
EӨcell = 0.00-(-0.76)
= + 0.76 V Using Nernst Equation: Ecell = EӨ
cell + 0.059/n lg [H+]2/[Zn2+]PH2
Ecell = 0.76 + 0.059/2 lg [H+]2 / (1)(1)
pH = (0.76 – Ecell ) / 0.059
If the emf for the electrochemical cell is + 0.54 V, the pH would be
pH = (0.76 – 0.54)/0.059 = 3.7
