Title: Lesson 4 Voltaic Cells
Learning Objectives: Explain in simple terms how voltaic cells use redox reactions to produce electricity Understand that oxidation occurs at the anode and reduction at the cathode Make a series of voltaic cells in order to better understand the how they work Refresh Consider the following three redox reactions.
Cd(s) + Ni2+(aq) Cd2+(aq) + Ni(s) Ni(s) + 2Ag+(aq) Ni2+(aq) + 2Ag(s) Zn(s) + Cd2+(aq) Zn2+(aq) + Cd(s) Deduce the order of reactivity of the four metals, cadmium, nickel, silver and zincand list in order of decreasing reactivity. Identify the best oxidizing agent and the best reducing agent. Electrochemical Cells
The fact that redox reactions involve transfers of electrons suggests a linkbetween this type of chemical reactivity and electricity. Voltaic Cells Voltaic cells generate electricity from spontaneous redox reactions Consider Zin reducing copper ions When the reaction is carried out in a single test tube, the electrons flowspontaneously from the zinc to the copper ions in the solution, energy is releasedin the form of heat (exothermic reaction) We can organize this reaction so that the energy is released in the form ofelectrical energy We need to separate the two half reactions Into half cells and allowing the electrons to flow between them only through anexternal circuit. This is a voltaic or a galvanic cell. Boardworks A2 Chemistry Redox Chemistry and Electrode Potentials
What is a half cell? If a rod of metal is dipped into a solution of its own ions, an equilibrium is set up. For example: Zn(s) Zn2+(aq)+2e- zinc metal strip zinc sulfate solution (1 mol dm-3) This is a half cell and the strip of metal is an electrode. The position of the equilibrium determines the potential difference between the metal strip and the solution of metal. Electrode potentials In a zinc half-cell, zinc atoms will form ions by releasing electronsthat will make the surface of the metal negatively charged withrespect to the solution There will therefore be a charge separation, known as anelectrode potential, between the metal and its ions in solution. At the same time, ions in the solutions gain electrons to form Zn atoms, so the equilibrium exists: The position of this equilibrium determines the size of the electrode potential in the half-cell, and depends on the reactivity of the metal. Because copper is the less reactive metal, in its half-cell the equilibrium position for the equivalent reaction lies further to the right: E.g. It has less tendency to lose electrons compared to zinc. Consequently, there are fewer electrons to the copper metal strip, so it will develop a higher (or less negative) electrode potential. Cells and electrode potentials
Boardworks A2 Chemistry Redox Chemistry and Electrode Potentials Cells and electrode potentials Two connected half-cells make a voltaic cell
If we connect these two half-cells by an external wire, electrons will have atendency to flow spontaneously from the zinc half-cell to the copper half-cellbecause of their different electrode potentials. The half-cells connected this way are called electrodes. The equilibrium for the copper half-cell lies further to the right than the equilibrium in the zinc half cell. ANODE OXIDATION -VE CHARGE CATHODE REDUCTION +VE CHARGE Key Parts of a Voltaic Cell
Anode Electrode or half-cell where oxidation happens Contains the more reactive metal The negative electrode: produces electrons Cathode Electrode or half-cell where reduction happens Contains the less reactive metal The positive electrode: accepts electrons External circuit Connects the metal electrodes in each half cell Electrons flow from anode to cathode Salt Bridge Contains a neutral salt such as potassium or sodium nitrate as it does not interfere with the reactions at theelectrodes Made of a tube of jelly or a filter paper soaked in salt solution Ions diffuse in and out to neutralise build up of charge, maintaining the potential difference. Voltmeter Measures the difference in potential between half-cells Could be replaced with other circuitry to do useful work REMEMBER AnOx Anode-Oxidation CaRe Cathode-Reduction Anions move in the salt bridge from the cathode to anode.
It opposes the flow of electrons in the external circuit. Cations move in the salt bridge from the anode to cathode. Without a salt bridge, no voltage is generated! Boardworks A2 Chemistry Redox Chemistry and Electrode Potentials
Combining half cells 1 Representing half cells: cell diagrams
Boardworks A2 Chemistry Redox Chemistry and Electrode Potentials Representing half cells: cell diagrams An electrochemical cell can be represented in a shorthand way by a cell diagram. E = V Zn(s) | Zn2+(aq) || Cu(aq) | Cu(s) E = V Anode Cathode The double vertical lines represents a salt bridge. The single lines represent a phase change between the solid metal and the aqueous metal ions. Aqueous solutions are placed next to the salt bridge. The half cell with the greatest negative potential is on the left of the salt bridge, so Ecell = Eright cell Eleft cell. In this case, Ecell = = V. The left cell is being oxidized while the right is being reduced. Different half-cells make voltaic cells with different voltages
Any two half-cells can be connected together to make a voltaic cell. The direction of electron flow and the voltage generated will be determined by thedifference in the reducing strength of the two metals. Can be judged by the relative position in the reactivity series. E.g.If we swap the copper half-cell for a silver half cell A larger voltage would be produced because the difference in electrode potentials will be greater. Electrons would flow from the zinc (anode) silver (cathode) If you now swap zinc for copper
Electrons would flow from copper(anode) to silver (cathode) Copper has a greater reducing power,it has the lower electrode potential. From these examples we can summarise: Electrons flow from anode to cathodethrough the external circuit Anions migrate from cathode to anodethrough the salt bridge Cations migrate from anode tocathode through the salt bridge Another example The reaction of Mg with Cu2+ ions:
Mg(s) + Cu2+(aq) Mg2+(aq) + Cu(s) This reaction involves two electrons being transferred from the Mg to the Cu: Mg Mg2+ + 2e- Cu2+ + 2e- Cu The Mg reduces the copper ions as it is more reactive This is an exothermic reaction, and the energy is normally released as heat A voltaic cell forces each half of the reaction to take place in a separate container, with theelectrons moving through a circuit to get from one side to the next This is an exothermic reaction, where the energy is released as electrical rather than thermalenergy The reactions in Voltaic cells usually involve only metals but do not have to. Voltaic Cells Continued
Electron Flow - + Electron Flow Anode: Whereoxidationhappens Cathode: Wherereductionhappens Constructing Voltaic Cells
You will need to build and measure the potential of voltaic cells comprisingvarious combinations of the following: Cu/Cu2+ Fe/Fe2+ Mg/Mg2+ Sn/Sn2+ Zn/Zn2+ Follow the instructions here General Reminders Comparisons of half-cell electrode potentials need a reference point
Potential difference is known as the electromotive force (EMF) Electrons tend to flow from half-cells: more negative potential more positive potential Potential generated is called the cell potential or electrode potential Symbol is E. Magnitude of this voltage depends on the difference in tendency of reduction of the half- cells. Cant measure an isolated half cell (no electron flow) So we measure against a fixed reference point STANDARD HYDROGENELECTRODE Standard Electrode Potential, Eo
Half Cell Standard Electrode Potential, Eo / V H+(aq) + e- H2(g) 0.00 Li+(aq) + e- Li(s) -3.04 Mn2+(aq)+ 2e- Mn(s) -1.19 Cu2+(aq)+ 2e- Cu(s) +0.34 Br2(l)+ e- Br-(aq) +1.07 This is the potential of astandard electroderelative to the standardhydrogen electrode. Always measure thepotential of the reduction Measured in Volts, V Full table in the databooklet Look at the table in the data booklet: What trends do you notice? How do the values relate to your ideasof reactivity? How do the values compare to thereactivity series you constructedearlier? Comparisons of half-cell electrode potentials need a reference point
Potential difference is known as the electromotive force (EMF) Electrons tend to flow from half-cells: more negative potential more positive potential Potential generated is called the cell potential or electrode potential Symbol is E. Magnitude of this voltage depends on the difference in tendency of reduction of the half-cells. Cant measure an isolated half cell (no electron flow) So we measure against a fixed reference point STANDARD HYDROGEN ELECTRODE The standard hydrogen electrode
Boardworks A2 Chemistry Redox Chemistry and Electrode Potentials The standard hydrogen electrode The Standard Hydrogen Electrode
Platinum is used as the conducting metal because it is fairly inert, will not ionize and can b a catalyst for reduction Surface of the metal is coated in finely divided platinum to increase surface area for absorption of hydrogen gas The electrode is bathed alternately in H2(g) and H+(aq), setting up an equilibrium Forward - Reduction Standard electrode potential defined as 0.00 V So we can measure and compare electrode potentials or other half-cells Backward - Oxidation Measuring Standard Electrode Potentials
Temperature = 298K Pressure = 100 kPa or 1 atm Substances must be pure If half cell does not include a solid metal, platinum is used Connecting wire Half-cells measured under these conditions are known as standard half-cells Metal Electrode Aqueous solution of metal ions: [M+] = 1.0 mol dm-3 Boardworks A2 Chemistry Redox Chemistry and Electrode Potentials
Combining half cells 2 Measuring Standard Electrode Potentials
When the standard hydrogen electrode is connected to another standard hall-cell, the EMF generated isknown as the Standard Electrode Potential. Given the symbol E +ve value for E (+0.34V) indicates that this has greater tendency to be reduced than H+ Electrons flow from hydrogen half-cell (oxidized) to copper half cell (reduced) Anode Cathode Overall reaction equation: More reactive metals tend to lose their electrons
H+ will bereduced Electrons flowtoward Hydrogen -ve value for E Anode OXIDISED Cathode REDUCED Overall reaction equation: From the examples we can see: Zinc has a lower E than hydrogen so can reduce H+ From the examples we can see: Copper has a higher E than hydrogen so cannot reduce H+ Remember, Standard Electrode Potentials are give for REDUCTION reaction
Sometimes known as the Standard Reduction Potentials Oxidised species on the left, reduced species on the right. Note: The E values do not depend on the total number of electrons, no need toscale according to stoichiometry The electrochemical series
Boardworks A2 Chemistry Redox Chemistry and Electrode Potentials The electrochemical series The electrochemical series is a list of standard electrode potentials (E). The equilibria are written with the electrons on the left of the arrow, i.e. as a reduction. E / V Half equation Half cell Mg2+(aq) / Mg(s) Mg2+(aq) + 2e Mg(s) -2.36 Zn2+(aq) / Zn(s) Zn2+(aq) + 2e Zn(s) -0.76 2H+(aq) / H2(g) 2H+(aq) + 2e H2(g) Cu2+(aq) / Cu(s) Cu2+(aq) + 2e Cu(s) +0.34 Teacher notes Note that this is an abbreviated version of the electrochemical series. Standard electrode potentials are sometimes called reduction potentials. Changing the conditions of an electrode, such as the concentration of ions or temperature, will change its electrical potential. Ag+(aq) / Ag(s) Ag+(aq) + e Ag(s) +0.80 Electrodes with negative values of E are better at releasing electrons (i.e. better reducing agents) than hydrogen. Boardworks A2 Chemistry Redox Chemistry and Electrode Potentials
Calculating Ecell The e.m.f of an electrochemical cell, Ecell, is the difference between the standard electrode potentials of the two half cells. Ecell=E (positive electrode)E (negative electrode) The positive electrode is taken to be the least negative half cell, and the negative electrode is the most negative half cell. This can be worked out from the electrode potentials values in the electrochemical series. Calculating Ecell: worked example
Boardworks A2 Chemistry Redox Chemistry and Electrode Potentials Calculating Ecell: worked example An electrochemical cell is set up using the two half reactions below. What potential difference Ecell would this cell generate? Zn2+(aq)+2e Zn(s) E = V Cu2+(aq)+2e Cu(s) E = V Ecell = E (positive electrode) E (negative electrode) The zinc half cell has the more negative potential and so forms the negative electrode. Therefore: Ecell = (+0.34) (-0.76) = V Calculating Ecell: combining half equations
Boardworks A2 Chemistry Redox Chemistry and Electrode Potentials Calculating Ecell: combining half equations To find the overall reaction occurring in the cell as a whole, the two half equations are added together: Cu2+(aq)+2e Cu(s) Because the zinc half cell forms the negative electrode of the cell, oxidation occurs at this electrode and the half equation must be reversed: Zn(s) Zn2+(aq)+2e- Teacher notes Students should be reminded that electrons are omitted from the final equation. The two half equations are added to give the overall cell reaction: Zn(s)+Cu2+(aq) Zn2+(aq)+Cu(s) Boardworks A2 Chemistry Redox Chemistry and Electrode Potentials
Calculating Ecell Remember, Standard Electrode Potentials are give for REDUCTION reaction
Sometimes known as the Standard Reduction Potentials Oxidised species on the left, reduced species on the right. The more positive the E value, the more readily it is reduced Note: The E values do not depend on the total number of electrons, no need to scale according tostoichiometry A selection of electrode potential values are given in section 24 of the IB data booklet