AP EAMCET · Chemistry · Electrochemistry
At 298 K, the following reaction takes place for a cell at the hydrogen electrode
\(\mathrm{H}^{+}(\mathrm{aq})+\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H}_2(1\) bar \()\)
The solution pH is 10.0. What is the hydrogen electrode potential in volts ?
\(\left(\frac{2.303 \mathrm{RT}}{\mathrm{~F}}=0.06 \mathrm{~V}\right)\)
- A \(-0.6\)
- B \(-0.06\)
- C \(+0.6\)
- D \(+0.06\)
Answer & Solution
Correct Answer
(A) \(-0.6\)
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