JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
What would be the electrode potential for the given half cell reaction at \(pH = 5\) ? ........... \(\mathrm{V}\) \(2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{O}_{2}+4 \mathrm{H}^{\oplus}+4 \mathrm{e}^{-} ; \mathrm{E}_{\mathrm{red}}^{0}=1.23 \mathrm{V}\) \((\mathrm{R}=8.314 \;\mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} ; \text { Temperature }=298 \;\mathrm{K} ;\) oxygen under std. atm. pressure of \(1 \;bar\))
- A \(1.52\)
- B \(2.56\)
- C \(0.36\)
- D \(3.56\)
Answer & Solution
Correct Answer
(A) \(1.52\)
Step-by-step Solution
Detailed explanation
\(\mathrm{O}_{2}(\mathrm{g})+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) ; \mathrm{E}_{\mathrm{red.}}^{0}=1.23 \mathrm{V}\) From nernst equation…
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