Question

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

• A. $ E_{H_2} = \frac{RT}{F} \times pH $
• B. $ E_{H_2} = \frac{RT}{F} \cdot\frac{1}{pH} $
• C. $ E_{H_2} = \frac{2.303 RT}{F} pH $
• D. $ E_{H_2} = -0.591 pH $

Answer 1

Answer: (D)

$2 H ^{+}+2 e^{-} \longrightarrow H _{2}$

According to Nernst equation,

$E =E^{\circ}+\frac{0.0591}{n} \log \frac{1}{\left[ H ^{+}\right]^{2}}$

$E =0-\frac{0.0591}{2} \log \left[ H ^{+}\right]^{2}$

$=-0.0591 pH$