Electrochemistry CH1121.

Electrochemistry CH1121

Redox Reactions Last semester we covered reactions with reduction (gain of electrons) and oxidation (loss of electrons) occurring in one equation For this unit we will have to split these equations into the oxidation half and reduction half to make them easier to use

Redox Reactions Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe2+(aq) Can be broken into half reactions: Oxidation: Sn2+(aq)  Sn4+(aq) + 2e- Reduction: 2Fe3+(aq) + 2e-  2Fe2+(aq) Same number of electrons lost and gained

Redox Reactions We can have these half reactions occurring in acidic or basic surroundings Must balance these equations in a special way depending on if it is an acidic or basic environment Add H+(aq) and H2O(l) in acidic environment Add OH-(aq) and H2O(l) in basic environment

Redox Reactions (Acidic)
Balancing in acidic solution Half reactions Balance half reactions Not O and H first Balance O by adding H2O(l) Balance H by adding H+(aq) Balance charge by adding e-’s Multiply to make electrons cancel Combine/cancel/check

Half reactions MnO4-(aq) + C2O42-(aq) Mn2+(aq) + CO2(aq) Becomes MnO4-(aq)  Mn2+(aq) C2O42-(aq)  CO2(g)

Balance anything that isn’t O and H MnO4-(aq)  Mn2+(aq) C2O42-(aq)  2CO2(g)

Balance O by adding H2O(l) MnO4-(aq)  Mn2+(aq) + 4H2O(l) C2O42-(aq)  2CO2(g)

Balance H by adding H+(aq) MnO4-(aq) + 8H+(aq)  Mn2+(aq) + 4H2O(l) C2O42-(aq)  2CO2(g)

Balance charge by adding e-’s 5e- + MnO4-(aq) + 8H+(aq)  Mn2+(aq) + 4H2O(l) C2O42-(aq)  2CO2(g) + 2e-

Multiply to make electrons cancel 2(5e- + MnO4-(aq) + 8H+(aq)  Mn2+(aq) + 4H2O(l)) 5(C2O42-(aq)  2CO2(g) + 2e-) Becomes: 10e- + 2MnO4-(aq) + 16H+(aq)  2Mn2+(aq) + 8H2O(l) 5C2O42-(aq)  10CO2(g) + 10e-

Combine/cancel/check 10e- + 2MnO4-(aq) + 16H+(aq) + 5C2O42-(aq)  2Mn2+(aq) + 8H2O(l) + 10CO2(g) + 10e- Becomes: 2MnO4-(aq) + 16H+(aq) + 5C2O42-(aq)  2Mn2+(aq) + 8H2O(l) + 10CO2(g)

Redox Reactions (Basic)
Balancing in basic solution Half reactions Balance half reactions Not O and H first Balance O by adding H2O(l) Balance H by adding H+(aq) Balance charge by adding e-’s Balance/neutralize H+(aq) by adding OH-(aq) to BOTH sides! Combine H+(aq) and OH-(aq) to make H2O(l) Multiply to make electrons cancel Combine/cancel/check

Half reactions CN-(aq) + MnO4-(aq)  CNO-(aq) + MnO2(s) Becomes: CN-(aq)  CNO-(aq) MnO4-(aq)  MnO2(s)

Balance anything that isn’t O and H CN-(aq)  CNO-(aq) MnO4-(aq)  MnO2(s)

Balance O by adding H2O(l) CN-(aq) + H2O(l)  CNO-(aq) MnO4-(aq)  MnO2(s) + 2H2O(l)

Balance H by adding H+(aq) CN-(aq) + H2O(l)  CNO-(aq) + 2H+(aq) MnO4-(aq) + 4H+(aq)  MnO2(s) + 2H2O(l)

Balance charge by adding e-’s CN-(aq) + H2O(l)  CNO-(aq) + 2H+(aq) + 2e- MnO4-(aq) + 4H+(aq) + 3e-  MnO2(s) + 2H2O(l)

Neutralize H+(aq) by adding OH-(aq) to both sides 2OH-(aq) + CN-(aq) + H2O(l)  CNO-(aq) + 2H+(aq) + 2e- + 2OH-(aq) 4OH-(aq) + MnO4-(aq) + 4H+(aq) + 3e-  MnO2(s) + 2H2O(l) + 4OH-(aq)

Combine H+(aq) and OH-(aq) to make H2O(l) 2OH-(aq) + CN-(aq) + H2O(l)  CNO-(aq) + 2e- + 2H2O(l) 4H2O(l) + MnO4-(aq) + 3e-  MnO2(s) + 2H2O(l) + 4OH-(aq)

Multiply to make e-’s cancel 3(2OH-(aq) + CN-(aq) + H2O(l)  CNO-(aq) + 2e- + 2H2O(l)) 2(4H2O(l) + MnO4-(aq) + 3e-  MnO2(s) + 2H2O(l) + 4OH-(aq))

Multiply to make e-’s cancel 6OH-(aq) + 3CN-(aq) + 3H2O(l)  3CNO-(aq) + 6e- + 6H2O(l) 8H2O(l) + 2MnO4-(aq) + 6e-  2MnO2(s) + 4H2O(l) + 8OH-(aq)

Combine/cancel/check 6OH-(aq) + 3CN-(aq) + 11H2O(l) + 2MnO4-(aq) + 6e-  3CNO-(aq) + 6e- + 10H2O(l) + 2MnO2(s) + 8OH-(aq) Becomes: 3CN-(aq) + H2O(l) + 2MnO4-(aq)  3CNO-(aq) + 2MnO2(s) + 2OH-(aq)

Redox Reactions Acidic and basic solutions are balanced the same way in the beginning Basic has the extra step of adding OH-(aq) to neutralize the H+(aq) ions Both types of solutions need to be combined, cancelled, and checked in the end

Electrochemical Cells
Voltaic or galvanic cells Spontaneous redox reactions are a source of electrical energy Electrochemical cells are used to house these energy producing reactions

2 solid metals connected by an external circuit are called electrodes Anode Electrode where oxidation takes place Cathode Electrode where reduction takes place

Half reactions can be used to represent what takes place at each electrode Anode (oxidation) Zn(s)  Zn2+(aq) + 2e- Cathode (reduction) Cu2+(aq) + 2e-  Cu(s)

Half-cells are electronically neutral As Zn is oxidized in anode cell, Zn2+ ions enter solution upsetting Zn2+/SO42- charge balance As Cu2+ is reduced in cathode cell, Cu ions enter colution upsetting Cu2+/SO42- charge balance

Salt bridge allows ions to migrate between cells to keep electronic neutrality Can also use a porous wall between cells Salt bridge is a U- shaped tube containing electrolyte solution that does not react with cell solutions

Anions always migrate toward the anode Cations always migrate toward the cathode SO42- ions move from the cathode to the anode Zn2+ ions move from the anode to the cathode

Electrons move from the anode to the cathode through an external circuit Due to the flow of electrons Anode = negative Cathode = positive

Standard Hydrogen Electrode
Has a reduction potential of 0V Used to compare and calculate the reduction potentials of all other substances Construction: H2(g) has pressure of 1atm H+(aq) has molarity of 1.0M 25℃ Gas bubbled over inert metal (usually platinum mesh) Voltage is measured

Standard Hydrogen Electrode
S.H.E

Gaseous Reactants Cells can be constructed just like the S.H.E. in order to find the cell potentials of any gaseous reactants

Standard Reduction Potentials
Tendency for a substance to be reduced More positive E⁰red more likely to be reduced Found by experiment only, given in a table (in your data booklet) Measured in volts (V) Standard oxidation and standard reduction potentials for a substance are of equal magnitude but opposite sign

Standard Reduction Potentials

Standard Cell Potential (E⁰cell)
We can use the values for E⁰red to determine the E⁰cell for a given electrochemical cell E⁰cell = E⁰red + E⁰ox Equation is NOT given! Coefficients do not effect voltage!

A voltaic (electrochemical) cell is based on the two standard half reactions: Cd2+(aq) + 2e-  Cd(s) Sn2+(aq) + 2e-  Sn(s) Determine which half-reaction occurs at the cathode and the anode Calculate the cell potential

E⁰red for Cd2+(aq) = -0.4 E⁰red for Sn2+(aq) = is more positive so Sn is reduced and Cd is oxidized Reduction occurs at cathode Oxidation occurs at anode Sn is cathode Cd is anode

Cd2+(aq) + 2e-  Cd(s) Becomes Cd(s)  Cd2+(aq) + 2e- And potential becomes +0.4

E⁰cell = E⁰red + E⁰ox E⁰cell = E⁰cell = +0.26V

Batteries Self contained electrochemical cells that can be used as a power source Primary: Not rechargeable Secondary Rechargeable

Dry Cell Batteries Primary battery Contain an electrolyte paste
Cheap to make but prone to leaking AAA, AA, C, and D Provide 1.5V

Alkaline Batteries Mostly primary, some secondary versions
Improved dry cell battery KOH added to electrolyte (alkaline) Fewer leaks, better performance in cold, last longer than dry cell

Lead-Acid Storage Batteries
Secondary battery Anode and cathode are made of lead components Entire contents are suspended in H2SO4(aq) Series of plates where lead is oxidized and reduced

Disadvantages Large and heavy Sulfuric acid and lead (toxic, corrosive, etc.) Overcharging can release hydrogen and oxygen gas and can be explosive

Advantages: Best performance of any battery Store large amounts of charge Deliver large currents Cost effective/efficient Hazards are manageable Recyclable

Nickel-Cadmium (NiCad) Batteries
Secondary battery Advantages Lasts many charging cycles Keep same power throughout use Disadvantages Cd is toxic (hard to recycle) Chemical memory means charging too soon decreases capacity Damaged by over charging Not as good as new alternatives

Fuel Cell Batteries Primary and secondary battery
Need supply of fuel (usually hydrocarbon) and oxygen Used in hybrid vehicles

Fuel Cell Batteries Advantages Disadvantages
Much more efficient compared to combustion Produce little to no pollution Disadvantages Safety issues with the use of hydrogen Expensive

Lithium Batteries Primary battery Contain a lithium anode
Current and reactivity of Li can be safety issue (explode/cause fires) Not the same as lithium ion and lithium polymer (cell phones, laptops, etc.)

Electrochemical Vs. Electrolytic Cells
Electrochemical Cell Electrolytic Cell Spontaneous Reaction Voltage is measured Anode negative Cathode positive Oxidation at anode Reduction at cathode Nonspontaneous Reaction External voltage applied Anode forced positive Cathode forced negative Oxidation at anode Reduction at cathode

Electrolytic Cell Electrolysis takes place in an electrolytic cell
Applying voltage to reverse the natural flow of electrons and ions Reduction is still at the cathode Oxidation is still at the anode

Electrolysis

Electrolysis With inert electrodes in a solution with many ions we may need to predict the products of electrolysis The redox reactions that occurs first will be the one that requires the least amount of voltage Go to “Standard Reduction Potentials” table Closest value to zero is the reaction to occurs first

Electrolysis A solution contains zinc chloride (ZnCl2) and iron (II) fluoride (FeF2). An external DC voltage is applied to inert platinum electrodes. Predict the products of the electrolysis.

Electrolysis The ions present in solution are Zn2+, Fe2+, Cl-, and F-.
In addition we need to consider that water itself can be oxidized or reduced if favourable.

Electrolysis Possible oxidations
2F-(aq)  F2(g) + 2e E° = -2.87V 2Cl-(aq)  Cl2(g) + 2e E° = -1.36V 2H2O(l)  O2(g) + 4H+(aq) + 4e- E° = -1.23V Since H2O(l) has a voltage closest to 0 it occurs first

Electrolysis Possible reductions
Fe2+(aq) + 2e-  Fe(s) E° = -0.44V Zn2+(aq) + 2e-  Zn(s) E° = -0.76V 2H2O(l) + 2e-  H2(g) + 2OH-(aq) E° = -0.83V Since Fe2+(aq) has a voltage closest to 0 it occurs first

Electrolysis For the overall reaction we have to combine the oxidation and reduction half reactions 2H2O(l)  O2(g) + 4H+(aq) + 4e- E° = -1.23V Fe2+(aq) + 2e-  Fe(s) E° = -0.44V Becomes Fe2+(aq) + 2e- + 2H2O(l)  Fe(s) + O2(g) + 4H+(aq) + 4e- E⁰cell = -1.23V V = -1.67V

Electrolysis For this electrolysis to happen we must apply slightly more than 1.67V Magnitude of E⁰cell tells you the magnitude of the voltage that must be applied

Electroplating Use of electrolysis to deposit a thin layer of one metal on another metal to improve the beauty or resistance to corrosion Metals get reduced at the cathode Cathode is the object being plated Examples Nickel of chromium onto steel Precious metal like silver onto less expensive Rhodium on jewelry (“dip”)

Faraday’s Law Faraday’s Law relates electric current, time, molar mass, and charge to mass of material deposited m = I · t · M ne · F I = current (amps) [C/s] t = time (s) M = molar mass (g/mol) ne = number of moles of electrons (coefficient of e- in equation) F = Faraday’s constant ( C/mol ) m = mass deposited

Faraday’s Law Calculate the number of grams of aluminum produced in 1.00h by the electrolysis of molten AlCl3 if the electrical current is 10.0A.

Looking at aluminum so deal with the half reaction:
Faraday’s Law Looking at aluminum so deal with the half reaction: Al3+ + 3e-  Al

Faraday’s Law 𝑚= 𝐼 𝑡 𝑀 𝑛 𝑒 𝐹 I = 10.0A t = 1.0hr = 3600s M = g/mol ne = 3 F = 96485C/mol

Faraday’s Law 𝑚= (10.0)(3600)( ) (3)(96485) =3.356𝑔𝐴𝑙 3.356g Al are deposited

Metallurgy The practice of engineering metals
Extraction Processing Preparing for practical uses Metals occur in nature as minerals Ores are mineral deposits from which metals can be produced economically

Metallurgy (Roasting)
Industrial process where an ore is heated in air to yield an oxide Oxides are easy to reduce to just the metal Examples: 2 ZnS(s) + 3 O2(g)  2 ZnO(s) + 2 SO2(g) 2 NiS(s) + 3 O2(g)  2 NiO(s) + 2 SO2(g) FeCO3(s)  FeO(s) + CO2(g)

Metallurgy (Pyrometallurgy)
Reduce a metal ore to the free metal Iron ore concentrate (FeO and Fe2O3), limestone (CaCO3) and coke (C) are added to the top of the blast furnace Carbon from the coke reacts with oxygen to form CO (reducing agent) CO reduces iron oxides to just iron

Metallurgy (Pyrometallurgy)

Metallurgy (Hall Process)
Bauxite a hydrate of aluminum oxide is found in nature Alumina (aluminum oxide) can be acquired through working with bauxite, this can then be combined with other agents to create pure aluminum Due to its unique characteristics aluminum required its own process to be refined

Metallurgy (Copper) Copper ore is first separated from waste by floatation Copper is then roasted to convert sulphides or carbonates to the oxide (easy to reduce) Copper oxides don’t need roasting Copper oxides are then heated to produce elemental copper

Metallurgy (Copper) Copper is then purified through electrolysis

Electrochemistry CH1121.

Similar presentations

Presentation on theme: "Electrochemistry CH1121."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Electrochemistry CH1121.

Similar presentations

Presentation on theme: "Electrochemistry CH1121."— Presentation transcript:

Similar presentations

About project

Feedback