Question

Calculate the reversible potential of oxygen electrode in a solution of $pH=1$, when the partial pressure of $O_2$ is ${10}^{-2}$atm [Assume $E_{O_2/H_{2}O}^0=1.23$V. vs SHE at $1$ atm and ${25}^o$C]

• A. $1.19V$
• B. $1.71V$
• C. $1.97V$
• D. $1.112V$

Answer 1

The correct option is C $1.97V$

$O_2+2H_{2}O+4e^{-}\to 4O{H}^{-}$

$pH=1$

$pOH=14-pH$

$pOH=14-1$

$pOH=13$

$\left[O{H}^{-}\right]={10}^{-pOH}$

$\left[O{H}^{-}\right]={10}^{-13}M$
$E_{O_2/H_{2}O}=E_{O_2/H_{2}O}^0-\frac{0.0591}{n}log\frac{[O{H}^{-}{]}^4}{P_{O_2}}$
$E_{O_2/H_{2}O}=1.23V-\frac{0.0591}{4}log\frac{({10}^{-13}{)}^4}{{10}^{-2}}$
$E_{O_2/H_{2}O}=1.23V-\frac{0.0591}{4}log\frac{({10}^{-13}{)}^4}{{10}^{-2}}$

$E_{O_2/H_{2}O}=1.97V$