Applications of Redox Electrochemistry

Review Oxidation reduction reactions involve a transfer of electrons. OIL- RIG Oxidation Involves Loss Reduction Involves Gain LEO-GER Lose Electrons Oxidation Gain Electrons Reduction

Solid lead(II) sulfide reacts with oxygen in the air at high temperatures to form lead(II) oxide and sulfur dioxide. Which substance is a reductant (reducing agent) and which is an oxidant (oxidizing agent)?   PbS, reductant; O2, oxidant  PbS, reductant; SO2, oxidant  Pb2+, reductant; S2- oxidant  PbS, reductant; no oxidant  PbS, oxidant; SO2, reductant

Applications Moving electrons is electric current. 8H++MnO4-+ 5Fe+2 +5e- ® Mn+2 + 5Fe+3 +4H2O Helps to break the reactions into half reactions. 8H++MnO4-+5e- ® Mn+2 +4H2O 5(Fe+2 ® Fe+3 + e- ) In the same mixture it happens without doing useful work, but if separate

e- e- e- e- e- H+ MnO4- Fe+2 Connected this way the reaction starts Stops immediately because charge builds up. e- e- e- e- e- H+ MnO4- Fe+2

Galvanic Cell Salt Bridge allows current to flow H+ MnO4- Fe+2

Electricity travels in a complete circuit H+ MnO4- Fe+2

Instead of a salt bridge Porous Disk H+ MnO4- Fe+2

e- e- e- e- Anode Cathode e- e- Reducing Agent Oxidizing Agent

Cell Potential Oxidizing agent pulls the electron. Reducing agent pushes the electron. The push or pull (“driving force”) is called the cell potential Ecell Also called the electromotive force (emf) Unit is the volt(V) = 1 joule of work/coulomb of charge Measured with a voltmeter

0.76 H2 in Cathode Anode H+ Cl- Zn+2 SO4-2 1 M ZnSO4 1 M HCl

Standard Hydrogen Electrode This is the reference all other oxidations are compared to Eº = 0 º indicates standard states of 25ºC, 1 atm, 1 M solutions. H2 in H+ Cl- 1 M HCl

Eºcell = EºZn® Zn+2 + EºCu+2 ® Cu Cell Potential Zn(s) + Cu+2 (aq) ® Zn+2(aq) + Cu(s) The total cell potential is the sum of the potential at each electrode. Eºcell = EºZn® Zn+2 + EºCu+2 ® Cu We can look up reduction potentials in a table. One of the reactions must be reversed, so change it sign.

Cell Potential Determine the cell potential for a galvanic cell based on the redox reaction. Cu(s) + Fe+3(aq) ® Cu+2(aq) + Fe+2(aq) Fe+3(aq) + e-® Fe+2(aq) Eº = 0.77 V Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V Cu(s) ® Cu+2(aq)+2e- Eº = -0.34 V 2Fe+3(aq) + 2e-® 2Fe+2(aq) Eº = 0.77 V

Reduction potential More negative Eº More positive Eº more easily electron is added More easily reduced Better oxidizing agent More positive Eº more easily electron is lost More easily oxidized Better reducing agent

Line Notation solid½Aqueous½½Aqueous½solid Anode on the left½½Cathode on the right Single line different phases. Double line porous disk or salt bridge. If all the substances on one side are aqueous, a platinum electrode is indicated.

For the last reaction Fe+2 Cu2+ Cu(s)½Cu+2(aq)½½Fe+2(aq),Fe+3(aq)½Pt(s) Cu2+ Fe+2

In a galvanic cell, the electrode that acts as a source of electrons to the solution is called the __________; the chemical change that occurs at this electrode is called________. a. cathode, oxidation b. anode, reduction c. anode, oxidation d. cathode, reduction

Under standard conditions, which of the following is the net reaction that occurs in the cell? Cd|Cd2+ || Cu2+|Cu a. Cu2+ + Cd → Cu + Cd2+ b. Cu + Cd → Cu2+ + Cd2+ c. Cu2+ + Cd2+ → Cu + Cd d. Cu + Cd 2+ → Cd + Cu2+

Galvanic Cell The reaction always runs spontaneously in the direction that produced a positive cell potential. Four things for a complete description. Cell Potential Direction of flow Designation of anode and cathode Nature of all the components- electrodes and ions

Practice Completely describe the galvanic cell based on the following half-reactions under standard conditions. MnO4- + 8 H+ +5e- ® Mn+2 + 4H2O Eº=1.51 V Fe+3 +3e- ® Fe(s) Eº=0.036V

Potential, Work and DG E = work done by system / charge E = -w/q emf = potential (V) = work (J) / Charge(C) E = work done by system / charge E = -w/q Charge is measured in coulombs. -w = q E Faraday = 96,485 C/mol e- q = nF = moles of e- x charge/mole e- w = -qE = -nFE = DG

Potential, Work and DG DGº = -nFEº if Eº > 0, then DGº < 0 spontaneous if Eº< 0, then DGº > 0 nonspontaneous In fact, reverse is spontaneous. Calculate DGº for the following reaction: Cu+2(aq)+ Fe(s) ® Cu(s)+ Fe+2(aq) Fe+2(aq) + e-® Fe(s) Eº = 0.44 V Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V

Cell Potential and Concentration Qualitatively - Can predict direction of change in E from LeChâtelier. 2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s) Predict if Ecell will be greater or less than Eºcell if [Al+3] = 1.5 M and [Mn+2] = 1.0 M if [Al+3] = 1.0 M and [Mn+2] = 1.5M if [Al+3] = 1.5 M and [Mn+2] = 1.5 M

E = Eº - RTln(Q) nF The Nernst Equation DG = DGº +RTln(Q) -nFE = -nFEº + RTln(Q) E = Eº - RTln(Q) nF 2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s) Eº = 0.48 V Always have to figure out n by balancing. If concentration can gives voltage, then from voltage we can tell concentration.

The Nernst Equation Eº = RTln(K) nF As reactions proceed concentrations of products increase and reactants decrease. Reach equilibrium where Q = K and Ecell = 0 0 = Eº - RTln(K) nF Eº = RTln(K) nF nF Eº = ln(K) RT

Batteries are Galvanic Cells Car batteries are lead storage batteries. Pb +PbO2 +H2SO4 ®PbSO4(s) +H2O

Batteries are Galvanic Cells Dry Cell Zn + NH4+ +MnO2 ® Zn+2 + NH3 + H2O + Mn2O3

Batteries are Galvanic Cells Alkaline Zn +MnO2 ® ZnO+ Mn2O3 (in base)

Batteries are Galvanic Cells NiCad NiO2 + Cd + 2H2O ® Cd(OH)2 +Ni(OH)2

Corrosion Rusting - spontaneous oxidation. Most structural metals have reduction potentials that are less positive than O2 . Fe ® Fe+2 +2e- Eº= 0.44 V O2 + 2H2O + 4e- ® 4OH- Eº= 0.40 V Fe+2 + O2 + H2O ® Fe2O3 + H+ Reactions happens in two places.

Salt speeds up process by increasing conductivity Water Rust Fe2+ O2 + 2H2O +4e- ® 4OH- e- Iron Dissolves- Fe ® Fe+2 Fe2+ + O2 + 2H2O ® Fe2O3 + 8 H+

Preventing Corrosion Coating to keep out air and water. Galvanizing - Putting on a zinc coat Has a lower reduction potential, so it is more easily oxidized. Alloying with metals that form oxide coats. Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.

Electrolysis Running a galvanic cell backwards. Put a voltage bigger than the potential and reverse the direction of the redox reaction. Used for electroplating.

1.10 e- e- Zn Cu 1.0 M Zn+2 1.0 M Cu+2 Anode Cathode

A battery >1.10V e- e- Zn Cu 1.0 M Zn+2 1.0 M Cu+2 Cathode Anode

Calculating plating Have to count charge. Measure current I (in amperes) 1 amp = 1 coulomb of charge per second q = I x t q/nF = moles of metal Mass of plated metal How long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag+

Calculating plating Current x time = charge Charge ∕Faraday = mole of e- Mol of e- to mole of element or compound Mole to grams of compound Or the reverse if you want time to plate

Calculate the mass of copper which can be deposited by the passage of 12.0 A for 25.0 min through a solution of copper(II) sulfate.

How long would it take to plate 5 How long would it take to plate 5.00 g Fe from an aqueous solution of Fe(NO3)3 at a current of 2.00 A?

Other uses Electrolysis of water. Separating mixtures of ions. More positive reduction potential means the reaction proceeds forward. We want the reverse. Most negative reduction potential is easiest to plate out of solution.

Redox 1. Know the table 2. Recognized by change in oxidation state. 3. “Added acid” 4. Use the reduction potential table on the front cover. 5. Redox can replace. (single replacement)

6. Combination Oxidizing agent of one element will react with the reducing agent of the same element to produce the free element. I- + IO3- + H+ ® I2 + H2O 7. Decomposition. a) peroxides to oxides b) Chlorates to chlorides c) Electrolysis into elements. d) carbonates to oxides

Examples A piece of solid bismuth is heated strongly in oxygen. A strip or copper metal is added to a concentrated solution of sulfuric acid. Dilute hydrochloric acid is added to a solution of potassium carbonate.

Hydrogen peroxide solution is added to a solution of iron (II) sulfate. Propanol is burned completely in air. A piece of lithium metal is dropped into a container of nitrogen gas. Chlorine gas is bubbled into a solution of potassium iodide.

Examples A stream of chlorine gas is passed through a solution of cold, dilute sodium hydroxide. A solution of tin ( II ) chloride is added to an acidified solution of potassium permanganate A solution of potassium iodide is added to an acidified solution of potassium dichromate.

Magnesium metal is burned in nitrogen gas. Lead foil is immersed in silver nitrate solution. Magnesium turnings are added to a solution of iron (III) chloride. Pellets of lead are dropped into hot sulfuric acid Powdered Iron is added to a solution of iron(III) sulfate.

A way to remember An Ox – anode is where oxidation occurs Red Cat – Reduction occurs at cathode Galvanic cell- spontaneous- anode is negative Electrolytic cell- voltage applied to make anode positive

Some AP Mult Choice Stats

A student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt. (a) Draw a diagram of this cell. (b) Describe what is happening at the cathode (Include any equations that may be useful.)

A student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt. (c) Describe what is happening at the anode. (Include any equations that may be useful.)

A student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt. (d) Write the balanced overall cell equation. (e) Write the standard cell notation.

A student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt. (f) The student adds 4 M ammonia to the copper sulfate solution, producing the complex ion Cu(NH3)+ (aq). The student remeasures the cell potential and discovers the voltage to be 0.88 volt. What is the Cu2+ (aq) concentration in the cell after the ammonia has been added?