Chemistry · Electrochemistry
JEE Main 2024 — 4 April, Shift 1 — Question 59
What pressure (bar) of would be required to make emf of hydrogen electrode zero in pure water at ?
- Option A:Correct
- Option B:
- Option C:
1
- Option D:
0.5
Answer: A
Step-by-step solution
\begin{array}{*{35}{r}}{} & \text{E}={{\text{E}}^{\text{o}}}-\frac{0.059}{\text{n}}\text{log}\frac{{{\text{P}}_{{{\text{H}}_{2}}}}}{{{\left[ {{\text{H}}^{+}}\right]}^{2}}}\\{}&0=0\frac{0.059}{2}\text{log}\frac{{{\text{P}}_{{{\text{H}}_{2}}}}}{{{\left({{10}^{-7}} \right)}^{2}}} \\{} & \text{log}\frac{{{\text{P}}_{{{\text{H}}_{2}}}}}{{{\left( {{10}^{-7}} \right)}^{2}}}=0 \\{} & \frac{{{\text{P}}_{{{\text{H}}_{2}}}}}{{{10}^{-14}}}=1 \\{} & {{\text{P}}_{{{\text{H}}_{2}}}}={{10}^{-14}}\text{bar} \\\end{array}
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2024
- Paper
- 4 April, Shift 1
- Subject
- Chemistry
- Chapter
- Electrochemistry
- Topic
- Nernst Equation and Electrochemical Series