AP EAMCET · Chemistry · Electrochemistry
The \(\mathrm{E}^{\circ}\) of \(\mathrm{M} \mid \mathrm{M}^{2+} \| \mathrm{Cu}^{2+} / \mathrm{Cu}\) is 0.3 V , At what concentration of \(\mathrm{Cu}^{2+}\) (in \(\mathrm{mol} \mathrm{L}^{-1}\) ), the \(\mathrm{E}_{\text {cell }}\) value becomes zero?
\(\left(\frac{2.303 R T}{F}=0.06\right)\left(\text { Conc. of } \mathrm{M}^{2+}=0.1 \mathrm{M}\right)\)
- A \(10^{-9}\)
- B \(10^{-8}\)
- C \(10^{-11}\)
- D \(10^{-10}\)
Answer & Solution
Correct Answer
(C) \(10^{-11}\)
Step-by-step Solution
Detailed explanation
According to Nernst equation, \(\mathrm{E}=\mathrm{E}^{\circ}-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \frac{[\text { product }]}{[\text { reactant }]}\)…
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