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Electrochemistry question

2013 · Q89
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Electrochemistry question

2013 · Q89

NEETChemistryElectrochemistryMCQ+4 / −1
A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be
  1. A
    0.118 V
  2. B
    1.18 V
  3. C
    0.059 V
  4. D
    0.59 V
View written solutionFree

Correct answer: D

Oxidation half reaction is

H2 →\to→ 2H+ + 2e–

If pH = 10, [H+] = 10–10

We know, Nernst Equation

Ecell = E°cell - 0.05912log⁡[H+]2PH2{{0.0591} \over 2}\log {{{{\left[ {{H^ + }} \right]}^2}} \over {{P_{{H_2}}}}}20.0591​logPH2​​[H+]2​

For hydrogen electrode E°cell = 0

Ecell = - 0.05912log⁡(10−10)21{{0.0591} \over 2}\log {{{{\left( {{{10}^{ - 10}}} \right)}^2}} \over 1}20.0591​log1(10−10)2​

= 0.0591log1010

Ecell = 0.591 V

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