Search any question & find its solution
Question:
Answered & Verified by Expert
The potential of hydrogen electrode of $\mathrm{pH}=10$ with respect to standard hydrogen electrode is
Options:
Solution:
1238 Upvotes
Verified Answer
The correct answer is:
$-0.591 \mathrm{~V}$
For, hydrogen electrode
$$
\mathrm{pH}=10 \quad\left[\mathrm{H}^{+}\right]=10^{-10} \mathrm{M} \quad \mathrm{H}^{+}+e^{-}=\frac{1}{2} \mathrm{H}_2
$$
$$
\begin{aligned}
& E_{\text {cell }}=E^{\circ}-\frac{0.0591}{1} \log \frac{\left[\mathrm{H}_2\right]}{\left[\mathrm{H}^{+}\right]} \\
& E^{\circ}=0 ; \log \mathrm{H}_2=0 \text { because } \\
& E_{\text {cell }}=-\frac{0.0591}{1} \log \left(\frac{1}{10^{-10}}\right)-0.0591 \times 10 \\
& E_{\text {cell }}=-0.591 \mathrm{~V}
\end{aligned}
$$
$$
\mathrm{pH}=10 \quad\left[\mathrm{H}^{+}\right]=10^{-10} \mathrm{M} \quad \mathrm{H}^{+}+e^{-}=\frac{1}{2} \mathrm{H}_2
$$
$$
\begin{aligned}
& E_{\text {cell }}=E^{\circ}-\frac{0.0591}{1} \log \frac{\left[\mathrm{H}_2\right]}{\left[\mathrm{H}^{+}\right]} \\
& E^{\circ}=0 ; \log \mathrm{H}_2=0 \text { because } \\
& E_{\text {cell }}=-\frac{0.0591}{1} \log \left(\frac{1}{10^{-10}}\right)-0.0591 \times 10 \\
& E_{\text {cell }}=-0.591 \mathrm{~V}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.