0 votes
in Chemistry Form 4 by
edited by

The table gives standard reduction potentials for some half cells.

 Half cell  Half cell equation   E0/V 
    I Fe3+(aq) + e → Fe2+(aq) +0.77 
   II K+(aq) +e → K(s) −2.92
   III Ag+(aq) + e → Ag(s) +0.80
   IV Pb2+(aq) +2e → Pb(s) −0.13
   V l2(aq) + 2e → 2l(aq) +0.54

  1. State the standard conditions of an electrochemical cell. 
  2. An electrochemical cell was constructed using half-cells III and IV.
    • Complete Figure 2 by labelling the parts of the cells indicated as A1- A4Electrochemical cell construction using half cells
      Figure 2
      • A1 
      • A2
      • A3
      • A4
    • Write an equation for the cell reaction and calculate the e.m.f. of the cell.
      • Equation      
      • e.m.f.      
    • The salt bridge helps in completing the circuit. Explain why a saturated solution of potassium chloride is not suitable for use in the salt bridge in this electrochemical cell.    
  3. State why it is not possible to construct a similar electrochemical cell using half-cells II and III. 
  4. State and explain the observations made when aqueous potassium iodide is added to aqueous iron(III) sulphate.  
  5. Acidified potassium dichromate(VI) and acidified potassium manganate(VII) may be used in determining concentration of Fe2+ ions in a sample. If acidified potassium dichromate(VI) is used, an indicator is added to determine the end point but for acidified potassium manganate(VII), no indicator is added.
    • Explain why it is not necessary to use an indicator when acidified potassium manganate(VII) is used.  
    • An alloy containing iron was dissolved in an acid and the total volume made up to 250cm3. 25.0 cm3 of this solution required 18.0 cm3 of 0.15 M acidified potassium dichromate(VI) to react completely. The equation for the reaction is:
      Cr2O72− (aq) + 6Fe2+(aq)+ 14H+(aq) →2Cr3+(aq)+ 6Fe3+(aq) +7H2O(l)
      Calculate the mass of iron in the alloy (Fe = 56.0).     

1 Answer

0 votes
    • 1M solution
    • 1 atmospheric pressure.
    • Temperature of 25°C or 278K
      Soluble salt of Pb 
      • A1 - Pb/Lead electrode 
      • A2 - Pb2+/ Lead(II)Nitrate solution/1M Pb2+ 
        Soluble salt of Ag   
      • A3 - Ag/ Silver electrode 
      • A4 - Ag+/ Silver Nitrate/1M Ag
      • Equation   
        Pb(s) + 2Ag+ (aq) → Pb2+ (aq) + 2Ag(s) 
      • e.m.f.    
    • Formation of insoluble PbCl2 That reduces the concentration of ions in electrolyte
      Formation of AgCl that reduces the effectiveness of the cell
  3. K reacts explosively with water.
    E.M.F of cell is very high which can explode with the cell.
    • (Yellow) Brown solutions turns to green.  Fe3+ ions are reduced to Fe2+
    • Grey/black precipitate is formed. Iodide ions are oxidised to iodine
      Fe3+(aq) + 2I(aq)      →  Fe2+(aq)   +    I2(s)              is +0.23V
      Brown/yellow solubles        Green       Grey/black        Explanation  
    • H+/KMnO4, acts as its own indicator changing from purple to colourless (decolourised)
    • Moles of Cr2O72−=(18 × 0.15) /1000   =0.0027
      Moles of Fe2+ in 25.0cm3 = 6 × 0.0027 = 0.0162
      Moles of Fe2+ in 250cm3 = 0.0162 × 10 = 0.162
              0.648M  in 250cm3
      M=(0.0162 × 1000)/25  =0.648M
      Mass of Iron=0.162 × 56
      Condensed working
      (18 × 0.15 × 6 × 250 × 56)      = 9.072g
Welcome to EasyElimu Questions and Answers, where you can ask questions and receive answers from other members of the community.

5.9k questions

8.3k answers


590 users