InterviewSolution
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InterviewSolution
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`Cu^(2+)+2e^(-) rarr Cu`. For this, graph between `E_(red)` versus `ln[Cu^(2+)]` is a straight line of intercept `0.34V`, then the electrode oxidation potential of the half cell `Cu|Cu^(2+)(0.1M)` will beA. `-0.34 + (0.0591)/(2)V`B. `0.34 + 0.0591 V`C. `0.34 V`D. None of these |
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Answer» Correct Answer - A `E_(Cu//Cu^(2+)) = E_(Cu//Cu^(2+))^(@) -(0.059)/(1)log[Cu^(2+)]` If `log [Cu^(2+)] = 0`, i.e., `[Cu^(2+)] = 1`, Then `E_(Cu//Cu^(2+)) = E_(Cu//Cu^(2+))^(@)` or Intercept `= E_(Cu^(2+)//Cu) = 0.34` Now `E_(Cu//Cu^(2+))^(@) = -0.34 - (0.059)/(2) log 0.1` `= -0.34 + (0.059)/(2) V` |
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