Search any question & find its solution
Question:
Answered & Verified by Expert
Calculate the potential of hydrogen electrode in contact with a solution whose \(\mathrm{pH}\) is 10 .
Solution:
1986 Upvotes
Verified Answer
For hydrogen electrode,
\(\mathrm{H}^{+}+\mathrm{e}^{-} \rightarrow \frac{1}{2} \mathrm{H}_2\)
Applying Nernst equation,
\(\mathrm{E}_{\mathrm{H}^{+}, \frac{1}{2} \mathrm{H}_2}=\mathrm{E}_{\mathrm{H}^{+} / \frac{1}{2} \mathrm{H}_2}^{\mathrm{n}}-\frac{0.0591}{\mathrm{n}} \log \frac{1}{\left[\mathrm{H}^{+}\right]}\)
\(=0-\frac{0.0591}{1} \log \frac{1}{10^{-10}}\left\{\begin{array}{c}\mathrm{pH}=10 \\ {\left[\mathrm{H}^{+}\right]=10^{-10} \mathrm{M}}\end{array}\right\}\)
\(=-0.0591 \times 10\)
\(=-0.591 \mathrm{~V}\)
\(\mathrm{H}^{+}+\mathrm{e}^{-} \rightarrow \frac{1}{2} \mathrm{H}_2\)
Applying Nernst equation,
\(\mathrm{E}_{\mathrm{H}^{+}, \frac{1}{2} \mathrm{H}_2}=\mathrm{E}_{\mathrm{H}^{+} / \frac{1}{2} \mathrm{H}_2}^{\mathrm{n}}-\frac{0.0591}{\mathrm{n}} \log \frac{1}{\left[\mathrm{H}^{+}\right]}\)
\(=0-\frac{0.0591}{1} \log \frac{1}{10^{-10}}\left\{\begin{array}{c}\mathrm{pH}=10 \\ {\left[\mathrm{H}^{+}\right]=10^{-10} \mathrm{M}}\end{array}\right\}\)
\(=-0.0591 \times 10\)
\(=-0.591 \mathrm{~V}\)
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.