Question

The standard electrode potential for the following reaction is $-0.57\text{V}$.
$Te{O}_3^{2-}(aq,1M)+3H_{2}O\left(l\right)+4e^{-}\to Te\left(s\right)+6O{H}^{-}\left(aq\right)$

• A. The potential at $pH=12.0$ is $-0.17\text{V}$
• B. The potential at $pH=12.0$ is $+0.21\text{V}$
• C. The potential at $pH=12.0$ is $-0.39\text{V}$
• D. The potential at $pH=12.0$ is $+1.95\text{V}$

Answer 1

The correct option is C The potential at $pH=12.0$ is $-0.39\text{V}$

$Te{O}_3^{2-}(aq,1M)+3H_{2}O\left(l\right)+4e^{-}\to Te\left(s\right)+6O{H}^{-}\left(aq\right)$

Here the number of electrons involved $=4$
We have $pH=12$
$\Rightarrow pOH=2$
$\Rightarrow \left[O{H}^{-}\right]={10}^{-2}$
$\therefore$ Applying the nernst equation, we have
$E=E^{\circ }-\frac{0.0591}{4}\times \text{log}\left(\frac{[O{H}^{-}{]}^6}{\left[Te{O}_3^{2-}\right]}\right)$
$=-0.57-\frac{0.0591}{4}\times \text{log}\left(\frac{({10}^{-2}{)}^6}{1}\right)$
$=-0.57-\frac{0.0591}{4}\times \text{log}\left({10}^{-12}\right)$
$=-0.57+0.1773$
$=-0.3927\text{V}$