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The pressure of H2 required to make the potential of H2-electrode zero in pure water at 298 K is:

  • Option 1)

    10-14 atm

  • Option 2)

    10-12 atm

  • Option 3)

    10-10 atm

  • Option 4)

    10-4 atm

 

Answers (1)

 

Electrode Potential(Nerst Equation) -

M^{n+}(aq)+ne^{-}\rightarrow M(s)

- wherein

E_{(M^{n+}/M)}=E_{(M^{n+}/M)}^{0}-\frac{RT}{nf}ln\frac{[M]}{[M^{n+}]}

[M^{n+}] is concentration of species

F= 96487 C moi^{-1}

R= 8.314 JK^{-1}moi^{-1}

T= Temperature in kelvin

 

 P (H) = 7 for water

-log [H^{+}] = 7

or -log [H^{+}] = 7\: or\: [H^{+}]

2 H^{+} + 2e^{-}\longrightarrow H_{2}(g)

E_{cell}=E^\circ\: cell -\frac{0.0591}{2} log \frac{PH_{2}^{-}}{[H^{+}]^{2}}

0=0-\frac{0.0591}{2} log \frac{PH_{2}}{(10^{-7})^{2}}

PH_{2} = 10^{-14} atm

 


Option 1)

10-14 atm

This is correct answer

Option 2)

10-12 atm

This is incorrect answer

Option 3)

10-10 atm

This is incorrect answer

Option 4)

10-4 atm

This is incorrect answer

Posted by

Vakul

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