TS EAMCET · Chemistry · Electrochemistry
What is the reduction electrode potential (in volts) of copper electrode when \(\left[\mathrm{Cu}^{2+}\right]=0.01 \mathrm{M}\) is in a solution at \(25^{\circ} \mathrm{C}\) ? \(\left(E^{\circ}\right.\) of \(\mathrm{Cu}^{2+} / \mathrm{Cu}\) electrode is \(\left.+0.34 \mathrm{~V}\right)\)
- A \(0.3991\)
- B \(0.2809\)
- C \(0.3105\)
- D \(0.3695\)
Answer & Solution
Correct Answer
(B) \(0.2809\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} E_{\text {cell }} & =E_{\text {cell }}^{\circ}+\frac{0.0591}{n} \log \left[M^{+}\right] \\ & =0.34+\frac{0.0591}{2} \log [0.01] \\ & =0.34-\frac{0.0591}{2} \times 2 \\ & =0.2809 \text { volt }\end{aligned}\)
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