TS EAMCET · Chemistry · Electrochemistry
At 298 K, if emf of the cell corresponding to the reaction, \(\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(0.01 \mathrm{M})+\mathrm{H}_2(\mathrm{~g})(1 \mathrm{~atm})\) is 0.28 V, then the pH of the solution at the hydrogen electrode is
\(\left(\frac{2.303 \mathrm{RT}}{\mathrm{~F}}=0.06 \mathrm{~V}\right),\left(\mathrm{E}_{\mathrm{Zn}^{2+} \mid \mathrm{Zn}}^{\mathrm{o}}=-0.76 \mathrm{~V}\right)\)
- A \(7\)
- B \(8\)
- C \(9\)
- D \(10\)
Answer & Solution
Correct Answer
(C) \(9\)
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