AP EAMCET · Chemistry · Electrochemistry
The minimum voltage (in \(\mathrm{V}\) ) required to bring about the electrolysis of \(1 \mathrm{M}\) copper (II) sulphate solution at \(298 \mathrm{~K}\) is
\[
\text { (Given } \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{H}_2 \mathrm{O} / \mathrm{H}^{+}}^{\circ}=-1.23 \mathrm{~V} \text { ) }
\]
- A 1.57
- B 0.89
- C -0.89
- D -1.57
Answer & Solution
Correct Answer
(B) 0.89
Step-by-step Solution
Detailed explanation
Oxidation : (anode) \[ 2 \mathrm{H}_2 \mathrm{O}(\mathrm{I}) \rightarrow \mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+} \text {(aq.) } ; \mathrm{E}^{\circ}=+1.23 \mathrm{~V} \] Reduction : (cathode)…
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