Question

The oxidation potential of a hydrogen electrode is related to the pH of the solution by the equation at 25${}^0$C :

• A. 0.059 x pH
• B. 0.059 - pH
• C. $\frac{1}{pH}$
• D. 0.059 + pH

Answer 1

Answer: (A)

0.059 x pH

Oxidation potential
$H\to H^{+}+e^{-}$
$E_{cell}^{\circ }=0$
$\therefore E_{cell}=0-0.059log\left[H^{+}\right]$
$=0.059\times pH$