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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
\[
\text { (Given } \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{H}_2 \mathrm{O} / \mathrm{H}^{+}}^{\circ}=-1.23 \mathrm{~V} \text { ) }
\]

  1. A 1.57
  2. B 0.89
  3. C -0.89
  4. D -1.57
Verified Solution

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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