Question

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- A₄
     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).     

Answer 1

1.   
   • 1M solution
   • 1 atmospheric pressure.
   • Temperature of 25°C or 278K
2.   
   •   
     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.    
       +0.8−−0.13 
       +0.93V
   • Formation of insoluble PbCl₂ That reduces the concentration of ions in electrolyte
     OR
     Formation of AgCl that reduces the effectiveness of the cell
3. K reacts explosively with water.
   OR
   E.M.F of cell is very high which can explode with the cell.
4.   
   • (Yellow) Brown solutions turns to green.  Fe³⁺ ions are reduced to Fe²⁺
   • Grey/black precipitate is formed. Iodide ions are oxidised to iodine
     Equation
     Fe³⁺(aq) + 2I⁻(aq)      →  Fe2⁺(aq)   +    I₂(s)              is +0.23V
     Brown/yellow solubles        Green       Grey/black        Explanation  
5.   
   • 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
                       =9.072g
     Condensed working
     (18 × 0.15 × 6 × 250 × 56)      = 9.072g
              1000×25