EMF of hydrogen electrode in term of pH is (at 1 atm pressure).

Question

EMF of hydrogen electrode in term of pH is (at 1 atm pressure). (A)  EH2 = (RT/F) × pH  (B)  EH2 = (RT/F) ⋅(1/pH)  (C)  EH2 = (2.303 RT/F) pH  (D)  EH2 = -0.591 pH

Tardigrade Admin · Accepted Answer

2 H ++2 e- longrightarrow H 2 According to Nernst equation, E =E°+(0.0591/n) log (1/[ H +]2) E =0-(0.0591/2) log [ H +]2 =-0.0591 pH

Q. EMF of hydrogen electrode in term of $p H$ is (at $1$ atm pressure).

$E_{H_{2}} = \frac{RT}{F} \times p H$

$E_{H_{2}} = \frac{RT}{F} \cdot \frac{1}{p H}$

$E_{H_{2}} = \frac{2.303 RT}{F} p H$

$E_{H_{2}} = - 0.591 p H$

$2 H^{+} + 2 e^{-} ⟶ H_{2}$

According to Nernst equation,

$E = E^{\circ} + \frac{0.0591}{n} lo g \frac{1}{[ H ^{+} ] ^{2}}$

$E = 0 - \frac{0.0591}{2} lo g [H^{+}]^{2}$

$= - 0.0591 p H$

Q. EMF of hydrogen electrode in term of pH is (at 1 atm pressure).

EH2​​=FRT​×pH

EH2​​=FRT​⋅pH1​

EH2​​=F2.303RT​pH

EH2​​=−0.591pH