ELECTROCHEMISTRY

What is Electrochemistry  The branch of chemistry that deals with the use of spontaneous chemical reaction to produce electricity and the use of electricity to bring about non spontaneous chemical change.

What are Half reactions?  Oxidation occurs together with redox reaction but it is helpful to consider each process separately by writing two half reactions.  Mg(s)  Mg 2+ (aq) + 2e -  Fe 3+ (aq) + 3e -  Fe(s)

Class Practice  Write the equation for the half reaction for (a) the oxidation of iron (II) ions to iron(III) ions in aqueous solution; (b) the reduction of copper (II) ions in aqueous solution to copper metal.

Balancing redox reactions  The chemical equation of a reduction half reaction is added to that of an oxidation half reaction to form the chemical equation for the overall redox reaction.  You can balance a redox reaction in acidic and basic reaction.

Class example  Permanganate ions, MnO 4 -, react with oxalic acid, H 2 C 2 O 4, in acidic aqueous solution to produce manganese(II)ions and carbon dioxide gas. The partial skeletal equation is MnO 4 - (aq) + H 2 C 2 O 4(aq)  Mn 2+ (aq) +CO 2(g)

In basic solution  The products of the reaction of bromide ions with permanganate ions in basic aqueous solution are solid manganese (IV) oxide, MnO 2, and bromate ions, BrO 3 -. Balance the chemical equation for the reaction.

Homework  Page 832  18.6,18.8

Galvanic cell The Galvanic cell, named after Luigi Galvani, consists of two different metals connected by a salt bridge or a porous disk between the individual half-cells. It is also known as a voltaic cell and an electrochemical cell.

What occurs in a galvanic cell:  The zinc electrode is losing mass as Zn metal is oxidized to Zn 2+ ions which go into solution.  The concentration of the Zn 2+ solution is increasing. Anions, negative ions (e.g. SO 4 2- ), are flowing from the salt bridge toward the anode to balance the positive charge of the Zn 2+ ions produced.

 The copper electrode is gaining mass as Cu 2+ ions in the solution are reduced to Cu metal.  The concentration of the Cu 2+ solution is decreasing.  Cations, positive ions (e.g. Na + ), are flowing from the salt bridge toward the cathode to replace the positive charge of the Cu 2+ ions that consumed.  A reaction may start at standard-state conditions, but as the reaction proceeds, the concentrations of the solutions change, the driving behind the reaction becomes weaker, and the cell potential eventually reaches zero.

 In a galvanic cell, a spontaneous chemical reaction tends to draw electrons into the cell through the cathode, the site of reduction, and to release them at the anode, the site of oxidation.

Notation For Cells  Instead of drawing a cell diagram, chemists have devised a shorthand way of completely describing a cell called line notation. This notation scheme places the constituents of the cathode on the right and the anode components on the left. For example, a half-cell containing 1M solutions of CuO and HCl and a Pt electrode for the reduction of Cu 2+ would be written as: Pt (s) | Cu 2+ (aq), H + (aq)

 Note that the spectator ions, oxide and chloride, have been left out of the notation and that the anode will be written to the far left.  The salt bridge or porous disk is shown in the notation as a double line ( || ). Therefore, a cell that undergoes the oxidation of magnesium by Al 3+ could have the following cell notation if the anode is magnesium and the cathode is aluminum:  Mg (s) | Mg 2+ (aq) || Al 3+ (aq) | Al (s)

 An electrode is designated by representing the interfaces between phases by Ι.  A cell diagram depicts the physical arrangement of species and inter faces with a salt bridge denoted by II.

Class Practice  Write the diagram for a cell that has a hydrogen electrode on the left, an iron (III)/iron(II) electrode on the right, and includes a salt bridge. Both electrode contacts are platinum.

Cell Potential  An idealized cell for the electrolysis of sodium chloride is shown in the figure below. A source of direct current is connected to a pair of inert electrodes immersed in molten sodium chloride. Because the salt has been heated until it melts, the Na + ions flow toward the negative electrode and the Cl - ions flow toward the positive electrode.  When Na + ions collide with the negative electrode, the battery carries a large enough potential to force these ions to pick up electrons to form sodium metal.

 Negative electrode (cathode): Na + + e --  Na  Cl - ions that collide with the positive electrode are oxidized to Cl 2 gas, which bubbles off at this electrode.  Positive electrode (anode): 2 Cl -  Cl e -  The net effect of passing an electric current through the molten salt in this cell is to decompose sodium chloride into its elements, sodium metal and chlorine gas.

 Electrolysis of NaCl:  Cathode (-): Na + + e -  Na  Anode (+): 2 Cl -  Cl e -  The potential required to oxidize Cl - ions to Cl 2 is volts and the potential needed to reduce Na + ions to sodium metal is volts. The battery used to drive this reaction must therefore have a potential of at least 4.07 volts.  This example explains why the process is called electrolysis. Electrolysis uses an electric current to split a compound into its elements.  Electrolysis: 2 NaCl(l)  2 Na(l) + Cl 2 (g)

Home work  Page 833  18.18,18.24,

Standard cell potential  A cell's standard state potential is the potential of the cell under standard state conditions, which is approximated with concentrations of 1 mole per liter (1 M) and pressures of 1 atmosphere at 25 o C.  To calculate the standard cell potential for a reaction:  Write the oxidation and reduction half-reactions for the cell.  Look up the reduction potential, E o reduction, for the reduction half-reaction in a table of reduction potentials

 Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For the oxidation half-reaction, E o oxidation = - E o reduction.  Add the potentials of the half-cells to get the overall standard cell potential.  E o cell = E o reduction + E o oxidation  Example:  Find the standard cell potential for an electrochemical cell with the following cell reaction.  Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s)

 Write the half-reactions for each process.  Zn(s)  Zn 2+ (aq) + 2 e - Cu 2+ (aq) + 2 e -  Cu(s)  Look up the standard potentials for the reduction half-reaction.  E o reduction of Cu2+ = V  Look up the standard reduction potential for the reverse of the oxidation reaction and change the sign.  E o reduction of Zn2+ = V E o oxidation of Zn = - ( V) = V  Add the cell potentials together to get the overall standard cell potential.  oxidation: Zn(s)  Zn 2+ (aq) + 2 e - E o ox. = - E o red. = - ( V) = V  reduction: Cu 2+ (aq) + 2 e -  Cu(s) E o red. = V overall: Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) E o cell = V

Class Practice  The standard potential of a Zn 2+ /Zn electrode is – 0.76 V, and the standard potential of the cell Zn(s)IZn 2+ (aq)IICu 2+ (aq)ICu(s) is 1.10 V. What is the standard potential of the Cu 2+ /Cu electrode?

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The Electrochemical Series  The oxidizing and reducing power of a substance can be determined by its position in the electrochemical series. The strongest oxidizing agents are at the top of the table as reactants, the strongest reducing agents are at the top of the table as products.

Class practice  Can aqueous potassium permanganate be used to oxidize iron(II) to iron(III) under standard conditions in acidic solution.

 To find a standard cell potential arising from a spontaneous reaction we must combine the standard potential of the cathode half reaction(reduction) with that of the anode half reaction (oxidation) in such a way so as to obtain a positive value. The overall potential must be positive because that corresponds to a spontaneous process, and only a spontaneous process can generate a potential. If the calculation results in a negative value, this means that the reverse reaction is spontaneous.

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Standard potential, free energy and equilibrium constant  The free energy change is a measure of the change in the total entropy of a system and its surroundings at constant pressure; spontaneous processes are accompanied by a decrease in free energy.  G=H −TS  H is the enthalpy  T is the temperature  S is the entropy

 When the reaction free energy ∆G f is negative, the cell reaction is spontaneous and the cell generates a positive potential. When ∆G f is is large as well as negative, the cell potential is high as well as positive. The relationship suggests that ∆G f = − nFE  n is the number of moles of electrons that are transferred between the electrodes for the cell reaction as written in the chemical equation.

 The Faraday constant, F is the magnitude of the charge per mole of electrons: F=N a e= X10 4 C/mol.  1C.V=1J  We can write F= X10 4 J/V.mol

Class Practice  The cell Cr(s) Ι Cr 2+ (aq) Ι Cu (s) was found to have E⁰ = +1.08V at 298K.(a) Write the balanced net equation for the cell reaction;(b) determine n; and© calculate the standard reaction free energy at 298K.

 The following cell was set up: Hg(l) ΙHg 2 Cl₂(s)ΙHCl (aq)ΙΙHg₂(NO₃)₂(aq)ΙHg(l), E⁰ =+ 0.52V at 298.(a) Write the equation for the cell reaction.(b) determine n, and © calculate the standard reaction free energy at 298K.

 The equilibrium constant of a reaction can be calculated from standard potentials by combining the equation for the half reactions to give the reaction of interest and determining the standard potential of the corresponding cell.

Class Practice  The reaction between zinc metal and iodine in water generates 1.30 v under standard conditions. Determine a)n and b) ∆G f ⁰ for the cell reaction Zn(s)+I 2(aq )  Zn 2+ (aq) + 2I - (aq)  The solubility product is the equilibrium constant for the dissolution of a salt. Calculate the solubility product of silver chloride.

The Nernst Equation  The variation of cell potential with composition is expressed by the Nernst equation:  E=E ⁰ ─(RT/nF) ln Q

Class Practice  Calculate the potential at 25 ⁰C of a Daniel cell in which the concentration of Zn² + ions is 0.10 mol/L and that of the Cu 2+ ions is mol/L

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Electrolysis  Electrolysis is a method of separating chemically bonded elements and compounds by passing an electric current through them.

 In an electrolytic cell, current supplied by an external source is used to drive a nonspontaneous redox reaction.  The Galvanic cell, consists of two different metals connected by a salt bridge or a porous disk between the individual half-cells. It is also known as a voltaic cell and an electrochemical cell.

The porous bridge is substituted by a Salt bridge for a galvanic cell.

 Oxidation-reduction or redox reactions take place in electrochemical cells. There are two types of electrochemical cells. Spontaneous reactions occur in galvanic (voltaic) cells; nonspontaneous reactions occur in electrolytic cells.  Both types of cells contain electrodes where the oxidation and reduction reactions occur. Oxidation occurs at the electrode termed the anode and reduction occurs at the electrode called the cathode.

 The anode of an electrolytic cell is positive (cathode is negative), since the anode attracts anions from the solution. However, the anode of a galvanic cell is negatively charged, since the spontaneous oxidation at the anode is the source of the cell's electrons or negative charge. The cathode of a galvanic cell is its positive terminal. In both galvanic and electrolytic cells, oxidation takes place at the anode and electrons flow from the anode to the cathode.

 The redox reaction in a galvanic cell is a spontaneous reaction. Hence, galvanic cells are commonly used as batteries. Galvanic cell reactions supply energy which is used to perform work. The energy is harnessed by situating the oxidation and reduction reactions in separate containers, joined by an apparatus that allows electrons to flow. A common galvanic cell is the Daniell cell.

Class practice  Predict the products resulting from the electrolysis of 1M ZnNO 2 (aq) at pH=7

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