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Electrochemistry question

2016 · Q89
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Electrochemistry question

2016 · Q89

NEETChemistryElectrochemistryMCQ+4 / −1
The pressure of H2 required to make the potential of H2-electrode zero in pure water at 298 K is
  1. A
    10−-−10 atm
  2. B
    10−-−4 atm
  3. C
    10−-−14 atm
  4. D
    10−-−12 atm
View written solutionFree

Correct answer: C

2H+ (aq)

  • 2e– →\to→ H2(g)

    Ecell = Eocell - 0.05912log⁡pH2[H+]2{{0.0591} \over 2}\log {{{p_{{H_2}}}} \over {{{\left[ {{H^ + }} \right]}^2}}}20.0591​log[H+]2pH2​​​

    ⇒\Rightarrow⇒ 0 = 0 - 0.05912log⁡pH2[10−7]2{{0.0591} \over 2}\log {{{p_{{H_2}}}} \over {{{\left[ {{{10}^{ - 7}}} \right]}^2}}}20.0591​log[10−7]2pH2​​​

    ⇒\Rightarrow⇒ pH2[10−7]2=1{{{p_{{H_2}}}} \over {{{\left[ {{{10}^{ - 7}}} \right]}^2}}} = 1[10−7]2pH2​​​=1

    ⇒\Rightarrow⇒ PH2 = 10–14 atm
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