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    Electrochemistry
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    Estimate the minimum potential difference needed to reduce Al2O3 at 500° C. The free energy change for the decomposition reaction.
    2Al2O3→ 43 Al+O2 is ∆G = +960 kJ.(F = 96500 C mol-1)
    Applying the formula:

    ∆G° = -nF°E
    960 × 103J = -4×E°×96500                      = -960 × 1034×96500 = E°cell              E°cell = -965003860=-2.487 V.
    Since,
              ∆G = +960 kJ = +960×103J,E° = ? F = 96500 C mol-1.43Al3++4e-  →  43Al  n = 4.
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    Answer
    The measured resistance of a conductance cell containing 7.5 x 10–3 M solution of KCl at 25° C was 1005 ohms. Calculate (a) specific conductance (b) molar conductance of the solution cell constant = 1.25 cm–1.
    We have given
    Resistance = 1005ohm
    cell constant =1.25 cm-1 
    thus
    Specific conductance,

    k = 1R×cell constant    = 11005×1.25 ⇒ 1.244×10-3S cm-1Λm = 1000 KM⇒ 1000×1.244×10-37.5×10-3S cm2 mol-1⇒  165 s cm2 mol-1.
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    Answer
    Calculate the number of coulombs required for the oxidation of 1 mole of water to oxygen as per equation:
    2H2O → 4H+ + O2 + 4e–
    [Given: 1 F = 96,500 C mol–1]

    2H2O → 4H+ + O2 + 4e–

    Formula required charge n × F
    n = difference of charge on ions   
    F is constant and equal to 96500 Coulombs
    Here n = 2


    2 moles of H2O require 4 x 96500 C
    1 mole of H2O will need

    = 4×965002 = 2×96500 = 193000 C

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    Answer
    The potential of a hydrogen electrode in a solution of unknown [H+] is 0.29 V at 298 k measured against a standard hydrogen electrode. Calculate the pH of the solution.
    The both electrode are same thus there Ecell0 =0
    Since the potential of the hydrogen electrode measured against standard hydrogen electrode is a positive value, the electrode is acting as the anode. If the unknown [H    +] be x, the Nernst equation takes the form
          Ecell = E°cell-0.05922logx212
      Since both electrodes are the same, E°cell = 0.
    ∴     0.29 V = 0-0.0592 V2× 2 log x
    or         0.29V = -0.0592 V × log x
       Since -logx = -logH+ = pH
                 0.029 V = 0.0592 V × pH
    or               pH = 0.290.0592=4.9.
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    Answer
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    Two students use same stock-solution of ZnSO4 and solution of CuSO4. The emf of one cell is 0.03 V higher than the other. The conc. of CuSO4 in the cell with higher emf value is 0.5 M. Find out the concentration of CuSO4 in the other cell (2.303 RT/F = 0.06).

    The cell may be represented as
    The cell may be represented as
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