An Energizing Experience

An Energizing Experience
Voltaic Cells aka Galvanic Cells Using the energy of a spontaneous redox reaction to do work. Chapter 20 An Energizing Experience

A piece of copper is dropped into an aqueous solution of zinc nitrate, and in another beaker, a piece of zinc is dropped into copper(II) nitrate. A reaction occurs in both situations A reaction occurs in only one of the situations A reaction does not occur in either situation.

Complete the “possible” reactions
A piece of copper dropped into aqueous zinc nitrate Cu(s) + Zn(NO3)2(aq) → A piece of zinc dropped into aqueous copper (II) nitrate Zn(s) + Cu(NO3)2(aq) →

Activity Series Increasing Activity You probably remember talking about the activity series earlier this unit.

A piece of copper is dropped into an aqueous solution of zinc nitrate, and in an other beaker a piece of zinc is dropped into copper(II) nitrate. A reaction occurs in both situations A reaction occurs in only one of the situations Since one metal is more active than the other A reaction does not occur in either situation.

Spontaneous or Not? A piece of copper dropped into aqueous zinc nitrate Cu(s) + Zn(NO3)2(aq) → No Reaction This reaction did NOT occur Cu is LESS active than Zn. This reaction is NOT spontaneous aka “thermodynamically unfavorable.” A piece of zinc dropped into aqueous copper (II) nitrate Zn(s) + Cu(NO3)2(aq) → Zn(NO3)2(aq) + Cu(s) This reaction DID occur Zn is more active than Cu. Zn has a greater potential to lose its electrons than Cu This reaction IS spontaneous aka “thermodynamically favorable.”

What type of reaction is this
What type of reaction is this? Zn(s) + Cu(NO3)2(aq) → Zn(NO3)2(aq) + Cu(s) Turn to your mate and discuss. Perhaps it may fit into more than one category.

This is a single replacement reaction Zn(s) + Cu(NO3)2(aq) → Zn(NO3)2(aq) + Cu(s)
It is also an oxidation reduction reaction What gains electrons during the reaction? Zn Zn2+ Cu Cu2+ NO3−

What type of reaction is this
What type of reaction is this? Zn(s) + Cu(NO3)2(aq) → Zn(NO3)2(aq) + Cu(s) This is a single replacement reaction. It is also an oxidation reduction reaction What gains electrons during the reaction? Zn Zn2+ Cu Cu2+ NO3−

Let’s turn this into a net ionic equation.
A piece of zinc dropped into aqueous copper(II) nitrate Zn(s) + Cu(NO3)2(aq) → Zn(NO3)2(aq) + Cu(s) This is a single replacement reaction. It is also an oxidation reduction reaction

Let’s write balanced net ionic equations for this reaction
A piece of zinc dropped into aqueous copper(II) nitrate Zn(s) + Cu(NO3)2(aq) → Zn(NO3)2(aq) + Cu(s) Zn + Cu2+ → Zn2+ + Cu Zn loses electrons = oxidation Cu2+ gains electrons = reduction How can I remember? LEO says GER OIL RIG Ole´

This single replacement, redox reaction is really made up of two half reactions, one oxidation, the other reduction. Net Ionic: Zn + Cu2+ → Zn Cu The reaction above, separated into two half reactions: Oxidation: Zn → Zn e− Reduction: Cu e− → Cu

We can separate these two reactions and force the electrons to transfer through an external wire.
The two half reactions each occur in their own compartment. Net Ionic: Zn + Cu2+ → Zn Cu Zn → Zn e− Cu e− → Cu We can use this natural potential of the movement of electrons to do work. By moving the electrons through an external channel instead of allowing the electrons to transfer directly. This is called a voltaic (or galvanic) cell, aka battery.

Electromotive Force We call the driving force that makes these reactions go the electrical potential. aka: electromotive force aka: emf symbolized as E Eº (naught) means at standard conditions (1 atm, 1 M) E is measured in volts 1 V = 1 J/coulomb One volt is the potential energy difference required to move one coulomb of charge (a “chunk” of charge). The voltmeter can measure that potential energy difference.

A schematic of the electrochemical cell
The two half reactions will occur each in their own compartment. Zn + Cu2+ → Zn Cu Note the voltage on the meter is reading zero 0.0 Write the half reaction that occurs in each compartment.

When the electrons start to flow, in what direction will they move through the wire?
Cu e− → Cu Zn → Zn e− 0.0 Left to right Right to left

When the electrons start to flow, in what direction will they move through the wire?
Cu e− → Cu Zn → Zn e− 0.0 Left to right Right to left Electrons need to move from the zinc (since zinc atoms are oxidizing) to the copper electrode so that the copper ions can be reduced.

Before anything happens, is the solution on the left charged or neutral?
The solution is positively charged. The solution is negatively charged. The solution is neutral. Cu e− → Cu Zn → Zn e− 0.0

Before anything happens, is the solution on the left charged or neutral?
The solution is positively charged. The solution is negatively charged. The solution is neutral. Remember is was Cu(NO3)2 that was dissolved. There will be a stoichiometric balance between Cu2+ and 2NO3− that are dissolved. Cu e− → Cu Zn → Zn e− 0.0 NO3− Cu2+ Zn2+

Keeping in mind what ions are in the left compartment: Cu2+ and 2 NO3− When the electrons start to flow, what will happen to the balance of charge in the left side solution? (the amount of + compared to −) Cu e− → Cu Zn → Zn e− 0.0 NO3− Cu2+ Zn2+ The solution will become more positive. The solution will become less positive (more negative) The charge balance will stay the same. I have no idea of how to decide.

The solution will become more positive.
Keeping in mind what ions are in the left compartment: Cu2+ and 2 NO3− When the electrons start to flow, what will happen to the balance of charge in the solution? (the amount of + compared to −) The solution will become more positive. The solution will become less positive (more negative) This is a fundamental problem, and will stop the reaction in its tracks, because the imbalance of charge will not allow any more electrons to move. Thus the voltage reads zero The charge balance will stay the same. Cu e− → Cu Zn → Zn e− 0.0 NO3− Cu2+ Zn2+

A schematic of the electrochemical cell
We need to provide a way for the charge balance to be maintained. We provide a salt bridge. Usually a non-reacting salt like KNO3 or NaNO3. As the left solution loses Cu2+ ions, thus less positive, K+ or Na+ ions flow into the left solution. As the right solution gets more Zn2+ ions in the solution, thus more positive, NO3− ions will flow into the right solution. Cu e− → Cu Zn → Zn e− NO3− Cu2+ Zn2+ Na+ NO3− 1.10

Let’s calculate the theoretical potential for the Zinc/Copper cell
Let’s calculate the theoretical potential for the Zinc/Copper cell. Using a standard reduction potential table. Use the reduction potential table to look up the Reduction Potential for each half reaction. Net Ionic: Zn + Cu+2 → Zn Cu Zn e− → Zn Eºred = −0.76 V Cu e− → Cu Eºred = V One half reaction is oxidized. Flip the reaction = change the sign Zn → Zn e− Eºox = V Add Eºred + Eºox = Eºcell

Electromotive Force for a Voltaic Cell
The Eº of the oxidation is best called Eºox To determine Eºoxidation you should reverse the sign of the Eºreduction for the substance that is oxidized. To determine the Ecell, simply add the Eºreduction and Eºoxidation of the two half reactions. Your text book will show this equation: Eºcell = Eºred - Eºred Standard potentials are intensive because the Eº measures the potential energy difference per electrical charge. They are NOT affected by the stoichiometry used to balance the redox equation. (of the reduction that occurs at the cathode) (of the oxidation that occurs at the anode)

Using emf to Predict Spontaneity
The more positive the standard reduction potential, the greater the tendency for that reduction half-reaction to occur. A voltaic cell will be spontaneous if Eºcell is positive.

Inspection of a Typical Voltaic Cell
Zn | Zn+2 || Cu+2 | Cu

Important Features and Vocabulary of the Voltaic Cell
One half-cell contains the chemicals that involve the oxidation half reaction. The other half-cell contains chemicals that involve the reduction half reaction. A solid electrical connector, called an electrode is placed in each half-cell compartment. This provides a means for the electrons to travel through the external pathway.

Electrodes The electrode may or may not actually participate in the reaction. If the electrode is participating (i.e. metal is a reactant), it is said to be an active electrode. In this reaction the Zn is an active electrode. Zn metal is actually reacting. Zn → Zn e− If the electrode is not participating, it is said to be an inert or passive electrode. In this reaction the Cu is a passive (inert) electrode. Reduction occurs at the cathode. (Red Cat) Oxidation occurs at the anode. (An Ox)

As the reaction proceeds
Na+ NO3− Cu e− → Cu Zn → Zn e− 1.10 NO3− Cu2+ Zn2+ The mass of both electrodes change. The mass of only one electrode changes. The mass of both electrodes stay the same.

During the running of the cell, where does the mass of the zinc electrode go?
Na+ NO3− Cu e− → Cu Zn → Zn e− 1.10 NO3− Cu2+ Zn2+ through the salt bridge and onto the Cu electrode into the solution as Zn2+ ions across the external wire as electrons to hook up with Cu2+ ions

Where does the mass of the zinc electrode go?
Na+ NO3− Cu e− → Cu Zn → Zn e− 1.10 NO3− Cu2+ Zn2+ through the salt bridge and onto the Cu electrode into the solution as Zn2+ ions across the external wire as electrons to hook up with Cu2+ ions

The mass of the copper electrode increases
Na+ NO3− Cu e− → Cu Zn → Zn e− 1.10 NO3− Cu2+ Zn2+ the same that the Zn electrode decreases. more than the Zn electrode decreases. less than the Zn electrode decreases. hmmm….it is impossible to know whether the mass change would be the same or different. Actually the mass of the electrodes do NOT change at all due to the law of conservation of matter.

The mass of the copper electrode increases
Na+ NO3− Cu e− → Cu Zn → Zn e− 1.10 NO3− Cu2+ Zn2+ the same that the Zn electrode decreases. more than the Zn electrode decreases less than the Zn electrode decreases hmmm….it is impossible to know whether the mass change would be the same or different. Actually the mass of the electrodes do NOT change at all due to the law of conservation of matter.

Na+ NO3− Cu e− → Cu Zn → Zn e− 1.10 NO3− Cu2+ Zn2+ During the reaction, the cathode increased by 0.64 g, 0.01 mol of Cu atoms are deposited on the cathode. 6.0 x 1021 electrons are transferred and the mass of the anode would decrease by 0.64 g 1.2 x 1022 electrons are transferred and the mass of the anode would decrease by 0.64 g 6.0 x 1021 electrons are transferred and the mass of the anode would decrease by 0.65 g 1.2 x 1022 electrons are transferred and the mass of the anode would decrease by 0.65 g

Na+ NO3− Cu e− → Cu Zn → Zn e− 1.10 NO3− Cu2+ Zn2+ During the reaction, the cathode increased by 0.64 g, mol of Cu atoms are deposited on the cathode. 6.0 x 1021 electrons are transferred and the mass of the anode would decrease by 0.64 g 1.2 x 1022 electrons are transferred and the mass of the anode would decrease by 0.64 g 6.0 x 1021 electrons are transferred and the mass of the anode would decrease by 0.65 g 1.2 x 1022 electrons are transferred and the mass of the anode would decrease by 0.65 g

The mass changes for the Cu/Zn cell are not different by very much because the molar masses of copper and zinc are very similar, and every time one ion of Cu is reduced, one atom of Zn is oxidized Cu e− → Cu Al → Al e− salt bridge V If the Zn electrode were replaced by Al, the molar masses are quite different and a different number atoms are oxidized / reduced. Write a net ionic balanced equation to represent this reaction.

Use your Reduction Potential Table to calculate the voltage of this cell.
Cu e− → Cu Al → Al e− salt bridge V

During the reaction, the cathode increased by 0
During the reaction, the cathode increased by 0.64 g, and the mass of the anode would decrease by Cu e− → Cu Al → Al e− salt bridge V 0.16 g 0.96 g 0.24 g 1.38 g 0.18 g 1.67 g 0.43 g 1.92 g 0.64 g

During the reaction, the cathode increased by 0
During the reaction, the cathode increased by 0.64 g, and the mass of the anode would decrease by Cu e− → Cu Al → Al e− salt bridge V g 0.43 g 0.015 g 0.64 g 0.18 g 0.96 g 0.27 g 0.40 g

You have electrodes of nickel, silver, aluminum and cobalt and a nitrate solution of each metal (Co2+, Ni2+). Write a balanced equation to represent the redox reaction that will produce the best (greatest) cell potential. e- anode to cathode RED CAT Mg = anode (-), Al = cathode (+)

You have electrodes of nickel, silver, aluminum and cobalt and a nitrate solution of each metal (Co2+, Ni2+). Write a balanced equation to represent the redox reaction that will produce the best cell potential. Ag+ + e− → Ag Eºred = V Ni2+ + 2e- → Ni Eºred = −0.25 V Co2+ + 2e- → Co Eºred = −0.28 V Al3+ + 3e- → Al Eºred = −1.66 V e- anode to cathode RED CAT Mg = anode (-), Al = cathode (+) The most negative reduction potential will be the metal that is oxidized (because the sign will get switched) and then the greatest positive reduction potential will be the metal that is reduced. Overall equation: Ag+ + Al  Al3+ + Ag

Write a balanced equation to represent the redox reaction that will produce the best cell potential: 3 Ag+ + Al  Al Ag Now draw a schematic diagram of the galvanic cell. Which direction do e- flow? Which is cathode? anode? Label the ions & metal electrodes of cathode and anode. What else must be done to make the cell functional? Which electrode gains mass? which loses? e- anode to cathode RED CAT Mg = anode (-), Al = cathode (+)

You have electrodes of nickel, silver, aluminum and cobalt and a nitrate solution of each metal. Write the reaction that will produce the best cell potential, then draw a schematic diagram of the galvanic cell. e− −> Which direction do e- flow? Which is cathode? anode? What else must be done to make the cell functional? need a salt bridge Which electrode gains mass? which loses? the Al anode loses mass, the Ag cathode gains mass salt bridge Al Ag e- anode to cathode RED CAT Mg = anode (-), Al = cathode (+) Al(NO3)3(aq) anode AgNO3(aq) cathode

SHE Standard Hydrogen Electrode
Find the only voltage on the reduction potential table with a value of zero.

How to Measure Ered for Half-cells?
We have seen how we can measure the potential of a cell (Ecell), but if reduction can’t take place without oxidation, how are the half-cell potentials (Ered) measured? Universally, the standard hydrogen electrode (SHE) has been arbitrarily assigned a half-cell potential (Ered) of zero.

Standard Hydrogen Electrode The reference for all other half-reaction potentials
47

Write the half reaction that occurs at a hydrogen electrode when it serves as the cathode of a voltaic cell. What is standard about the Standard Hydrogen Electrode? What is the role of platinum foil in a standard hydrogen electrode? 48

Relative Reduction Potentials?
Since reduction only occurs while oxidation occurs at the same time, it is impossible to measure one without the other. To determine half reaction values, some half reaction had to be set as 0 The Standard Hydrogen Electrode (SHE) has been chosen as 0 1atm H2 1 M H+ H2 → 2H e− 2H e− → H2

Why the platinum electrode? It has a high level of inertness (it does not corrode). It has a high capability to catalyze the reaction of proton reduction because of its high intrinsic exchange density for protons It has had very high reproducibility of the potential 1atm H2 1 M H+ H2 → 2H e− 2H e− → H2

When connected to a zinc/zinc nitrate half cell, the potential for the complete cell is 0.76 V We know that the Zn is the anode. By defining the reduction of H+ as 0.0, we conclude that the oxidation of Zn is 0.76 V Thus the reduction of Zn2+ will be V

When connected to a copper/copper nitrate half cell, the potential for the complete cell is 0.34 V This time we know that the copper is the cathode By defining the reduction of H+ as 0.0, we conclude that the reduction of Cu2+ is 0.34 V By connecting many different half cells together, all the potentials can be determined relative to each other.

Mg(s) + Al3+(aq) → Mg2+(aq) + Al(s) Write out the aluminum half-reaction and determine the voltage, Eº for the half reaction you write. Volts = work/charge or J/coulomb

Mg(s) + Al3+(aq) → Mg2+(aq) + Al(s) Write out the aluminum half-reaction and determine the voltage, Eº for the half reaction you write. Al3+ + 3e− → Al Eºred = −1.66 V (J/coulomb) Volts = work/charge or J/coulomb

Mg(s) + Al3+(aq) → Mg2+(aq) + Al(s) Write out the magnesium half-reaction and determine the voltage, Eº for the half reaction you write.

Mg → Mg2+ + 2 e− Eºred = -2.37 V (J/coulomb)
Mg(s) + Al3+(aq) → Mg2+(aq) + Al(s) Write out the magnesium half-reaction and determine the voltage, Eº for the half reaction you write. Eºred = V (J/coulomb) Mg → Mg e− Eºox = V (J/coulomb)

Balance the equation Mg(s) + Al+3(aq) → Mg+2(aq) + Al(s)

Balance the equation Mg(s) + Al3+(aq) → Mg2+(aq) + Al(s)

What is the Eºcell 3Mg(s) + 2Al+3(aq) → 3Mg+2(aq) + 2Al(s)
Al3+ + 3e− → Al Eºred = −1.66 V (J/coulomb) Mg → Mg e− Eºox = V (J/coulomb) Eºred + Eºox = Eºcell − = Eºcell = 0.71 V (J/coulomb) This is an intensive property – the coefficients don’t affect the E° of the cell. Eºcell = 0.71

Draw a sketch of a Galvanic cell at standard conditions to represent this reaction. 3Mg(s) + 2Al3+(aq) → 3Mg2+(aq) + 2Al(s) Start with a generalized sketch of a Galvanic cell set-up. e- anode to cathode RED CAT Mg = anode (-), Al = cathode (+)

Draw a sketch of a Galvanic cell at standard conditions to represent this reaction. 3Mg(s) + 2Al3+(aq) → 3Mg2+(aq) + 2Al(s) Now think about the following details: Which direction do e- flow? Which is cathode? anode? What else must be done to make the cell functional? Which electrode gains mass? which loses? e- anode to cathode RED CAT Mg = anode (-), Al = cathode (+)

Draw a Galvanic cell at standard conditions to represent this reaction
Draw a Galvanic cell at standard conditions to represent this reaction. 3Mg(s) + 2Al3+(aq) → 3Mg2+(aq) + 2Al(s) ←← e- salt bridge Al Mg Which direction do e- flow? Which is cathode? anode? What else must be done to make the cell functional? need a salt bridge bridge Which electrode gains mass? which loses? the Mg anode loses mass, the Al cathode gains mass Al(NO3)3(aq) cathode Mg(NO3)2(aq) anode e- anode to cathode RED CAT Mg = anode (-), Al = cathode (+)

You have copper, iron, and lead metals and their respective M2+ nitrate solutions. What combination would make the cell with the highest voltage? Identify the possible half reactions. Determine the potential of the cell. Write the balanced net ionic equation for the cell. Which metal is the cathode? Anode? Complete the sketch with solutions and labels on the electrodes. Show the direction that electrons flow through the wire?

You have copper, iron, and lead metal
You have copper, iron, and lead metal. What combination would make the cell with the highest voltage? Cu+2 + 2e- → Cu Eºred = 0.34 V Pb+2 + 2e- → Pb Eºred = V Fe+2 + 2e- → Fe Eºred = -.44 V Identify the two half reactions. Determine the potential of the cell. Write the balanced net ionic equation for the cell. Which metal is the cathode? Anode? Complete the sketch with solutions and labels on the electrodes. Show the direction that electrons flow through the wire?

You have copper, iron, and lead metal
You have copper, iron, and lead metal. What combination would make the cell with the highest voltage? Cu(NO3)2(aq) Fe(NO3)2(aq) Cu Fe ←← e- Cu+2 + 2e- → Cu Eºred = 0.34 V Pb+2 + 2e- → Pb Eºred = V Fe+2 + 2e- → Fe Eºred = -.44 V Identify the two half reactions. Determine the potential of the cell. Write the balanced net ionic equation for the cell. Which metal is the cathode? Anode? Complete the sketch with solutions and labels on the electrodes. Show the direction that electrons flow through the wire?

You have copper, iron, and lead metals and their respective M2+ nitrate solutions. What combination would make the cell with the highest voltage? Cu(NO3)2(aq) Fe(NO3)2(aq) Cu Fe ←← e- Cu+2 + 2e- → Cu Eºred = 0.34 V Pb+2 + 2e- → Pb Eºred = V Fe+2 + 2e- → Fe Eºred = -.44 V Identify the two half reactions. the oxidation of iron and reduction of copper Determine the potential of the cell. Eºcell = 0.44 V V = 0.78 V Write the balanced net ionic equation for the cell. Fe + Cu+2 → Cu + Fe+2 Which metal is the cathode? Anode? RED CAT: copper is the cathode, iron is the anode. Complete the sketch with solutions and labels on the electrodes. Show the direction that electrons flow through the wire? electrons flow from anode to cathode

Hydrogen Peroxide

Write the balanced molecular equation for the decomposition of hydrogen peroxide.
2 H2O2 (aq)  2 H2O (l) + O2 (g) Is this a redox reaction? Write out the oxidation #s to find out. Yes. The oxygen gets reduced & oxidized. What’s the name for this again? Disproportionation

Now watch as your teacher performs this decomposition reaction with some added substances (some soap to make it foamier & a catalyst that makes it happen faster).

Write out the reduction half-reaction
Write out the reduction half-reaction. Use your blue paper and find the E°red for this half reaction. H2O2 + 2H+ + 2e− → 2H2O Eºred = V

The catalyst was a sodium-halide compound
The catalyst was a sodium-halide compound. We want to figure out which one. Write the half-reaction for the oxidation of all 4 halide ions without using your reduction potential table. Record the oxidation potential as well (you can use your blue sheet for this – remember you want the OXIDATION potential). Note that you look up the reduction and revers it/change sign for oxidation.

2F− → F2 + 2e− Eºox = −2.87 V 2Cl− → Cl2 + 2e− Eºox = −1.36 V
Write the half-reaction for the oxidation of all 4 halide ions. Record the oxidation potential as well. 2F− → F2 + 2e− Eºox = −2.87 V 2Cl− → Cl2 + 2e− Eºox = −1.36 V 2Br− → Br2 + 2e− Eºox = −1.07 V 2I− → I2 + 2e− Eºox = −0.53 V Note that you look up the reduction and revers it/change sign for oxidation.

Review of what the oxidation potentials tell us: Which halide is easiest to oxidize?
2F− → F2 + 2e− Eºox = −2.87 2Cl− → Cl2 + 2e− Eºox = −1.36 2Br− → Br2 + 2e− Eºox = −1.07 2I− → I2 + 2e− Eºox = −0.53

Which halide is easiest to oxidize?
2F− → F2 + 2e− Eºox = −2.87 2Cl− → Cl2 + 2e− Eºox = −1.36 2Br− → Br2 + 2e− Eºox = −1.07 2I− → I2 + 2e− Eºox = −0.53 Since the least negative value (or most positive value) would be most easily oxidized, the iodide is the most easily oxidized. What would be the diatomic (of these 4) that is most easily REDUCED?

2F− → F2 + 2e− Eºox = −2.87 2Cl− → Cl2 + 2e− Eºox = −1.36
H2O2 + 2H+ + 2e− → 2H2O Eºred = Which halide ion(s) should be able to be oxidized by hydrogen peroxide? Select all that apply. 2F− → F2 + 2e− Eºox = −2.87 2Cl− → Cl2 + 2e− Eºox = −1.36 2Br− → Br2 + 2e− Eºox = −1.07 2I− → I2 + 2e− Eºox = −0.53 75

H2O2 + 2H+ + 2e− → 2H2O Eºred = Which halide ion(s) should be able to be oxidized by hydrogen peroxide? Let’s try it. 2F− → F2 + 2e− Eºox = −2.87 2Cl− → Cl2 + 2e− Eºox = −1.36 2Br− → Br2 + 2e− Eºox = −1.07 2I− → I2 + 2e− Eºox = −0.53 76

The Cl− and Br− were unable to start the reduction, hmmmm
The Cl− and Br− were unable to start the reduction, hmmmm. What about the oxidation half- reaction? Write it out and find the E°ox of the oxidation half-reaction.

The Cl− and Br− were unable to start the reduction, hmmmm.
H2O2 → O2 + 2H+ + 2e− Eºox = −0.68 V H2O2 + 2H+ + 2e− → 2H2O Eºred = V

Reconsider the decomposition of hydrogen peroxide in light of the peroxide’s ability to be both reduced and oxidized. Which halide ions can actually be oxidized during this decompostion? H2O2 → O2 + 2H+ + 2e- Eºox = -0.68 H2O2 + 2H+ + 2e- → 2H2O Eºred = 2F− → F2 + 2e− Eºox = −2.87 2Cl− → Cl2 + 2e− Eºox = −1.36 2Br− → Br2 + 2e− Eºox = −1.07 2I− → I2 + 2e− Eºox = −0.53 Only the I-1 ion. While the peroxide reduction has enough emf to drive Cl-1 and Br-1, its own oxygen would oxidize first, since it has a lower Eºox

Reconsider the oxidation of halide ions by hydrogen peroxide. Which halide ions can actually be oxidized by H2O2? H2O2 → O2 + 2H+ + 2e− Eºox = -0.68 H2O2 + 2H+ + 2e− → 2H2O Eºred = 2F− → F2 + 2e− Eºox = −2.87 2Cl− → Cl2 + 2e− Eºox = −1.36 2Br− → Br2 + 2e− Eºox = −1.07 2I− → I2 + 2e− Eºox = −0.53 Only the I-1 ion. While the peroxide reduction has enough emf to drive Cl-1 and Br-1, its own oxygen would oxidize first, since it has a lower Eºox

The oxygen in hydrogen peroxide is both oxidized and reduced when hydrogen peroxide decomposes.
Write the overall reaction. 2H2O2 → 2H2O + O2 Eº = V Write the reduction half reaction. H2O2 + 2H+ + 2e− → 2H2O Eºred = 1.78 V Write the oxidation half reaction. H2O2 → O2 + 2H+ + 2e− Eºox = −0.68 V The best catalyst must be NaI since its oxidation potential easily allows both parts of the reaction to occur.

The Effect of Concentration on Voltage

Calculate the voltage of the following cell
1 M Cu(NO3)2 Cu salt bridge +0.34 V -0.34 V 0.0 V impossible to determine

the voltage of the two half reactions cancel out impossible to determine 0.0 1 M Cu(NO3)2 Cu salt bridge Cu → Cu e− Cu e− → Cu Cu → Cu e− Cu e− → Cu −−−−−−−−−−−−−− no reaction

greater than V more than V but less than 0.0 V less than 0.34 V but greater than 0.0 V more negative than V (that is to say, smaller than -0.34V) V 0.001 M Cu(NO3)2 5 M Cu(NO3)2 Cu salt bridge

0.0 0.001 M Cu(NO3)2 5 M Cu(NO3)2 Cu +0.34 V -0.34 V 0.0 V greater than V more than V but less than 0.0 V less than 0.34 V but greater than 0.0 V more negative than V (that is to say, smaller than -0.34V) This is called a Concentration Cell The voltage will be greater in magnitude than 0.34 or -0.34, the sign depends on how the voltmeter is hooked up.

greater than V less than 1.10 V but greater than 0.0 V more than V but less than 0.0 V more negative than V (that is to say, smaller than V) Cu e− → Cu Zn → Zn e− 0.0 2 M Cu(NO3)2 0.01 M Zn(NO3)2

The voltage for this cell would be greater than 1.10 V
Zn + Cu2+ → Zn Cu Since the concentration of the reactant is greater than standard condition, and since the concentration of the product is less than standard conditions we can predict that the reaction has an even greater tendency to occur in the forward direction and is therefore greater than V Cu e− → Cu Zn → Zn e− V 2 M Cu(NO3)2 0.01 M Zn(NO3)2

The Nernst Equation The mathematics that supports the calculation of nonstandard
Zn + Cu2+ → Zn Cu The mathematics that supports the calculation of the voltage for nonstandard conditions. Cu e− → Cu Zn → Zn e− 0.0 2 M Cu(NO3)2 0.01 M Zn(NO3)2

that’s it for now.....

An Energizing Experience

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