CHAPTER 15 Electro-chemistry 15.4 Electrochemical Cells

The lemon battery Now we know it is an oxidation reaction A chemical reaction between magnesium and the citric acid generates electrons Now we know it is an oxidation reaction Electron flow Magnesium Citric acid

The lemon battery Now we know it is an oxidation reaction A chemical reaction between magnesium and the citric acid generates electrons Now we know it is an oxidation reaction This means that somewhere else, a reduction reaction must also take place to absorb the electrons LED Electron flow Copper Citric acid

Electron flow Electron flow LED Oxidation and reduction reactions are separated Electron flow Electron flow Copper Magnesium Citric acid electrochemical cell: a device in which redox reactions take place.

Three main components of an electrochemical cell: 1) Two electrodes 2) The electrolyte 3) A conducting path that connects the electrodes externally electrochemical cell: a device in which redox reactions take place.

An electrochemical cell Three main components of an electrochemical cell: 1) Two electrodes The anode electrode is where the oxidation occurs The cathode electrode is where the reduction occurs

An electrochemical cell Three main components of an electrochemical cell: 1) Two electrodes 2) The electrolyte Electrodes are immersed in a conductive solution. This solution, or electrolyte, contains free ions.

An electrochemical cell Three main components of an electrochemical cell: 1) Two electrodes 2) The electrolyte 3) A conducting path that connects the electrodes externally There must be a connection between oxidation and reduction Electrons must be able to flow from the anode to the cathode

There are two types of electrochemical cells: Voltaic cells Spontaneous chemical reactions at the electrodes generate electrical current (A battery is a voltaic cell) Electrolytic cells An external electric current drives nonspontaneous reactions between the electrodes and the electrolyte (or galvanic cells)

There are two types of electrochemical cells: Voltaic cells Spontaneous chemical reactions at the electrodes generate electrical current (A battery is a voltaic cell) Electrolytic cells An external electric current drives nonspontaneous reactions between the electrodes and the electrolyte (or galvanic cells) In both cases, the cells are constructed with two half-cells: Anode where oxidation happens Cathode where reduction happens

A voltaic cell The flow of electrical current is completed inside the battery Negative ions move from the cathode to the anode Positive ions move from the anode to the cathode to maintain electroneutrality inside the battery

Chemistry in a voltaic cell Consider the following setup: Mg loses electrons more easily than Cu so electrons would flow from the Mg side e– flow e– flow Oxidation (anode): Mg(s) → Mg2+(aq) + 2e– Reduction (cathode): Cu2+(aq) + 2e– → Cu(s) Mg(s) Cu(s)

Chemistry in a voltaic cell Consider the following setup: e– flow e– flow Why does the flow of electrons only happen for a short time? Mg(s) Cu(s) LED no longer lights up

Chemistry in a voltaic cell Consider the following setup: Oxidation (anode): Mg(s) → Mg2+(aq) + 2e– e– flow e– flow Mg(s) Cu(s) Mg2+ ions have nowhere to go Mg2+ and e– are attracted to each other The reaction stops as soon as it starts

Chemistry in a voltaic cell Oxidation (anode): Mg(s) → Mg2+(aq) + 2e– Reduction (cathode) Cu2+(aq) + 2e– → Cu(s) e– flow e– flow If e– flows away, there is an excess of positive ions (Mg2+) If e– flows in, Cu2+ is consumed, and there is an excess of negative ions (SO42–) Mg(s) Cu(s)

Chemistry in a voltaic cell The solution: provide a path for positive ions to move from the anode half-cell to the cathode half-cell e– flow e– flow If e– flows away, there is an excess of positive ions (Mg2+) If e– flows in, Cu2+ is consumed, and there is an excess of negative ions (SO42–) Mg(s) Cu(s)

salt bridge: an electrical connection between the oxidation and the reduction half-cells of an electrochemical cell.

Notation

Electromotive force Electrons that flow from the anode to the cathode are “pushed” by the electromotive force electromotive force (emf): the difference in the electrical potential between the anode and the cathode of an electrochemical cell.

Faraday constant = 96,500 C/mole Electromotive force Based on the emf, we can calculate the amount of energy that the cell can provide: cell emf (the driving force of the cell reaction) electrical energy output moles of electrons Faraday constant = 96,500 C/mole electromotive force (emf): the difference in the electrical potential between the anode and the cathode of an electrochemical cell.

Electromotive force The Mg–Cu cell has a cell emf of Ecell = 2.71 V. The cell reaction is: Calculate the total energy that can be obtained from this cell if 2.3 g of Mg is consumed in the reaction. Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)

Electromotive force Half-reactions: The Mg–Cu cell has a cell emf of Ecell = 2.71 V. The cell reaction is: Calculate the total energy that can be obtained from this cell if 2.43 g of Mg is consumed in the reaction. Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s) Asked: Wcell, energy from this cell with 2.43 g of Mg Given: The cell reaction, Ecell = 2.71 V, and the amount of reactant Relationships: Half-reactions: Mg(s) → Mg2+(aq) + 2e– Cu2+(aq) + 2e– → Cu(s)

Electromotive force Half-reactions: Asked: Wcell, energy from this cell with 2.43 g of Mg Given: The cell reaction, Ecell = 2.71 V, and the amount of reactant Relationships: Half-reactions: Solve: Mg(s) → Mg2+(aq) + 2e– Cu2+(aq) + 2e– → Cu(s) From the oxidation half-reaction, 1 mole Mg ~ 2 moles e– So n = 0.0200 moles e–

Electromotive force Half-reactions: Asked: Wcell, energy from this cell with 2.43 g of Mg Given: The cell reaction, Ecell = 2.71 V, and the amount of reactant Relationships: Half-reactions: Solve: Answer: With 2.43 g of Mg, this cell will release 5,230 J of energy. Mg(s) → Mg2+(aq) + 2e– Cu2+(aq) + 2e– → Cu(s) From the oxidation half-reaction, 1 mole Mg ~ 2 moles e– So n = 0.0200 moles e–

Ecell = Ereduction + Eoxidation The total cell voltage (Ecell) is a combination of the potential at each half-cell: Ecell = reduction potential + oxidation potential Ecell = Ereduction + Eoxidation For the oxidation half-reaction: Mg(s) → Mg2+(aq) + 2e– Eoxidation = +2.37 V For the reduction half-reaction: Mg2+(aq) + 2e– → Mg(s) Ereduction = –2.37 V

Scientists decided to only keep track of Ereduction The total cell voltage (Ecell) is a combination of the potential at each half-cell: Ecell = reduction potential + oxidation potential Ecell = Ereduction + Eoxidation For the oxidation half-reaction: Mg(s) → Mg2+(aq) + 2e– Eoxidation = +2.37 V For the reduction half-reaction: Mg2+(aq) + 2e– → Mg(s) Ereduction = –2.37 V Scientists decided to only keep track of Ereduction

Standard reduction potentials Mg2+(aq) + 2e– → Mg(s) Ereduction = –2.37 V Standard conditions: Temperature = 25oC Pressure = 1 atm (for gases) Concentration = 1 M (for solutions) The cell voltage under standard conditions is called the standard reduction potential, Eocell

Standard reduction potentials Ecell = Ereduction + Eoxidation We can’t measure the potential of single electrodes We need a reference half-cell!

Standard reduction potentials Ecell = Ereduction + Eoxidation We can’t measure the potential of single electrodes We need a reference half-cell! Use this reaction as the reference where Ereduction = 0 V

Standard reduction potentials The potential of all other cells is measured with respect to the reference cell Reference half-cell

Using the standard reduction potentials, calculate the cell voltage (Eocell) of the cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s).

Using the standard reduction potentials, calculate the cell voltage (Eocell) of the cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s). Asked: Eocell of the reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Given: The cell reaction, standard reduction potentials Relationships: Eocell = Eoreduction + Eooxidation Look up Table 15.4

Zn(s) → Zn2+(aq) + 2e– Eooxidation = +0.76 V Using the standard reduction potentials, calculate the cell voltage (Eocell) of the cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s). Asked: Eocell of the reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Given: The cell reaction, standard reduction potentials Relationships: Eocell = Eoreduction + Eooxidation Zn(s) → Zn2+(aq) + 2e– Eooxidation = +0.76 V Cu2+(aq) + 2e– → Cu(s) Eoreduction = +0.34 V reverse reaction From the Eoreduction table

Zn(s) → Zn2+(aq) + 2e– Eooxidation = +0.76 V Using the standard reduction potentials, calculate the cell voltage (Eocell) of the cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s). Asked: Eocell of the reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Given: The cell reaction, standard reduction potentials Relationships: Eocell = Eoreduction + Eooxidation Zn(s) → Zn2+(aq) + 2e– Eooxidation = +0.76 V Cu2+(aq) + 2e– → Cu(s) Eoreduction = +0.34 V Solve: Eocell = Eoreduction + Eooxidation Eocell = 0.34 V + 0.76 V = 1.10 V Answer: The cell voltage is 1.10 V

Reaction spontaneity Eocell = Eoreduction + Eooxidation The sign of Eocell also tells you whether the reaction is spontaneous or nonspontaneous Eocell > 0 Spontaneous reaction Eocell < 0 Nonspontaneous reaction

Determine whether the reaction Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s) is spontaneous under standard conditions.

The half-cells and their half-cell potentials: Determine whether the reaction Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s) is spontaneous under standard conditions. Asked: The spontaneity of the reaction under standard conditions Given: The reaction and the table of standard reduction potentials Relationships: Eocell = Eoreduction + Eooxidation The half-cells and their half-cell potentials: Zn(s) → Zn2+(aq) + 2e– Eooxidation = +0.76 V Ni2+(aq) + 2e– → Ni(s) Eoreduction = –0.23 V

The half-cells and their half-cell potentials: Determine whether the reaction Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s) is spontaneous under standard conditions. Asked: The spontaneity of the reaction under standard conditions Given: The reaction and the table of standard reduction potentials Relationships: Eocell = Eoreduction + Eooxidation The half-cells and their half-cell potentials: Zn(s) → Zn2+(aq) + 2e– Eooxidation = +0.76 V Ni2+(aq) + 2e– → Ni(s) Eoreduction = –0.23 V Solve: Eocell = –0.23 V + 0.76 V = +0.53 V > 0 Answer: Since the total cell voltage is a positive number the reaction is spontaneous and proceeds as indicated.

Three main components of an electrochemical cell: 1) Two electrodes 2) The electrolyte 3) A conducting path that connects the electrodes externally (like a salt bridge)

Faraday constant = 96,500 C/mole cell emf (the driving force of the cell reaction) electrical energy output moles of electrons Faraday constant = 96,500 C/mole Eocell = Eoreduction + Eooxidation Standard conditions: Temperature = 25oC Pressure = 1 atm (for gases) Concentration = 1 M (for solutions)

standard cell potential Nernst equation Batteries (a voltaic cell) run out of energy Nernst equation: standard cell potential moles of electrons Under standard conditions: Q = 1 so that Ecell = Eocell

Nernst equation Batteries (a voltaic cell) run out of energy

The nonspontaneous reaction that is induced is called electrolysis Electrolytic cells There are two types of electrochemical cells: Voltaic cells Spontaneous chemical reactions at the electrodes generate electrical current (A battery is a voltaic cell) Electrolytic cells An external electric current drives nonspontaneous reactions between the electrodes and the electrolyte (or galvanic cells) The nonspontaneous reaction that is induced is called electrolysis

Electrolytic cells There are two types of electrochemical cells: Voltaic cells Spontaneous chemical reactions at the electrodes generate electrical current (A battery is a voltaic cell) (or galvanic cells) When we use a battery it acts as a voltaic cell

Electrolytic cells There are two types of electrochemical cells: Voltaic cells Spontaneous chemical reactions at the electrodes generate electrical current (A battery is a voltaic cell) Electrolytic cells An external electric current drives nonspontaneous reactions between the electrodes and the electrolyte (or galvanic cells) When we recharge a battery it acts as an electrolytic cell

Electrolytic cells Decomposition of water: 2H2O(l) → 2H2(g) + O2(g)

Electrolytic cells Decomposition of water: 2H2O(l) → 2H2(g) + O2(g) +1 –2 0 0 oxidation numbers reduction oxidation

nonspontaneous reaction Electrolytic cells Decomposition of water: 2H2O(l) → 2H2(g) + O2(g) +1 –2 0 0 oxidation numbers reduction oxidation Reduction: 2H2O(l) + 2e– → H2(g) + 2OH–(aq) Eored = –0.83 V Oxidation: 2H2O → O2(g) + 4H+(aq) + 4e– Eoox = –1.23 V Eocell = –2.06 V nonspontaneous reaction

nonspontaneous reaction Electrolytic cells Decomposition of water: 2H2O(l) → 2H2(g) + O2(g) +1 –2 0 0 oxidation numbers reduction oxidation Hydrogen (H2) is being considered as a source of energy for the future. The production of hydrogen from water is the cleanest but also the most expensive way of producing hydrogen. Eocell = –2.06 V nonspontaneous reaction