Chemical Cells

Chemical Energy  Heat Energy When magnesium powder is added into copper(II) sulphate solution, the temperature of the mixture rises. Displacement reaction occurs. Chemical Energy  Heat Energy

Overall equation (Redox reaction): Mg(s) + CuSO 4 (aq)  MgSO 4 (aq) + Cu(s) Ionic equation: Mg(s) + Cu 2+ (aq)  Mg 2+ (aq) + Cu(s) Half equations: Mg(s)  Mg 2+ (aq) + 2e - Oxidation Cu 2+ (aq) + 2e -  Cu(s) Reduction

Chemical Energy  Heat Energy Observable changes: mass of the magnesium strip decreases copper deposits on the copper strip blue colour of solution fades out.

Chemical Energy  Electrical Energy When electrons flow through an external circuit, a simple chemical cell is formed. Chemical Energy  Electrical Energy

Chemical Cell of Mg/Cu couple Magnesium is more reactive than copper, it oxidizes and loses electrons to form positive ions more readily. Mg(s)  Mg 2+ (aq) + 2e - Magnesium ions dissolve into the solution. Thus the mass of magnesium strip decreases.

Chemical Cell of Mg/Cu couple Electrons flow through the external circuit to the copper strip. Voltmeter shows positive deflection.

Chemical Cell of Mg/Cu couple Copper(II) ions in electrolyte move to the copper strip. They reduce and gain electrons to form copper atoms. Cu 2+ (aq) + 2e -  Cu(s) Thus copper deposits on the copper strip.

Simple chemical cell Basic requirements: Two different metals dipped in a solution of electrolyte

Simple chemical cell The more reactive metal forms ions more readily. It oxidizes and loses electrons. It becomes the negative electrode. The less reactive metal becomes the positive electrode. Positive metal ions dissolves into the electrolyte and electrons flow from the more reactive metal through the external circuit to the less reactive metal.

Simple chemical cell The voltage of the cell gives a measure of how strongly the electrons are ‘pushed’ through the circuit, and is measured by a voltmeter. The voltmeter should be correctly connected. The negative terminal should be connected to the more reactive metal while the positive terminal to the less reactive metal. Positive ions in electrolyte move to the less reactive metal. They reduce and gain electrons to form metal.

Comparing the tendency to form ions of different metals The more reactive metal should be connected to the negative terminal of voltmeter.

Comparing the tendency to form ions of different metals The greater the difference in their tendencies to form ions, the higher is the voltage of the cell.

Chemical cell of Cu/Ag couple Electrolyte is in the form of filter paper soaked with sodium chloride solution.

Chemical cell of Cu/Ag couple Copper is more reactive than silver, it oxidizes and loses electrons to form positive ions more readily. Cu(s)  Cu 2+ (aq) + 2e - Copper(II) ions dissolve into the electrolyte. Thus the mass of copper strip decreases. Copper is the negative electrode.

Chemical cell of Cu/Ag couple Electrons flow from copper through the external circuit to silver. Silver is the positive electrode. Hydrogen ions in electrolyte move to the silver electrode. They reduce and gain electrons to form hydrogen gas. 2H + (aq) + 2e -  H 2 (g) Thus effervescence occurs at silver strip. Overall reaction: Cu(s) + 2H + (aq)  Cu 2+ (aq) + H 2 (g)

The electrochemical series of metals Metals arranged in order of their tendencies to form ions. The order of metals in the electrochemical series is the same as that in their reactivity series (except for the position of calcium).

The electrochemical series of metals

Modification of simple chemical cell Two different metals dipped in two separate electrolytes. The two electrolytes are connected by a salt bridge which can be made by soaking a piece of filter paper in saturated potassium nitrate solution.

Modification of simple chemical cell

Anode (oxidation): Mg(s)  Mg 2+ (aq) + 2e - Cathode(reduction): Cu 2+ (aq) + 2e -  Cu(s) Overall equation: Mg(s) + Cu 2+ (aq)  Mg 2+ (aq) + Cu(s)

Modification of simple chemical cell

Salt bridge Two main functions: It completes the circuit by allowing ions to move towards one half cell from the other. It provides ions to balance the charges in the solutions of the two half cells. Salt bridge must not be dried. .

Chemical Energy  Heat Energy When excess FeSO 4 (aq) is added into a purple solution of acidified KMnO 4 (aq), the colour changes to yellow. acidified potassium permanganate solution iron(II) sulphate solution

Chemical Energy  Heat Energy Oxidation half equation:  Fe 2+ (aq)  Fe 3+ (aq) + e   Green yellow Reduction half equation:  MnO 4 – (aq) + 8H + (aq) + 5e –  Mn 2+ (aq) + 4H 2 O( )  Purple colourless Overall equation: 5Fe 2+ (aq) + MnO 4 – (aq) + 8H + (aq)  5Fe 3+ (aq) + Mn 2+ (aq) + 4H 2 O( )

Other forms of chemical cells: inert electrodes

Anode (oxidation):  Fe 2+ (aq)  Fe 3+ (aq) + e  Cathode(reduction):  MnO 4 – (aq) + 8H + (aq) + 5e –  Mn 2+ (aq) + 4H 2 O( ) Overall equation: 5Fe 2+ (aq) + MnO 4 – (aq) + 8H + (aq)  5Fe 3+ (aq) + Mn 2+ (aq) + 4H 2 O( )

Other forms of chemical cells: inert electrodes Observable changes: Green iron(II) sulphate solution changes to yellow Purple potassium permanganate solution changes to colourless (purple colour fades.)

Chemical Energy  Heat Energy When excess colourless KI(aq) is added into a yellow solution of Fe 2 (SO 4 ) 3 (aq), the colour changes to brown. iron(III) sulphate solution potassium iodide solution

Chemical Energy  Heat Energy Anode (oxidation):  2I - (aq)  I 2 (aq) + 2e   Colourless brown in KI Cathode(reduction): Fe 3+ (aq) + e   Fe 2+ (aq) yellow green Overall equation: 2Fe 3+ (aq) + 2I – (aq)  2Fe 2+ (aq) + I 2 (aq)

Other forms of chemical cells: inert electrodes

Anode (oxidation):  2I - (aq)  I 2 (aq) + 2e  Cathode(reduction): Fe 3+ (aq) + e   Fe 2+ (aq) Overall equation: 2Fe 3+ (aq) + 2I – (aq)  2Fe 2+ (aq) + I 2 (aq)

Other forms of chemical cells: inert electrodes Observable changes: Yellow iron(III) sulphate solution changes to green Colourless potassium iodide solution changes to brown because iodine formed will combine with potassium iodide to form a brown compound. I 2 (aq) + KI(aq)  KI 3 (aq)