TS EAMCET · Chemistry · Electrochemistry
A solution of \(\mathrm{Fe}^{2+}\) is titrated potentiometrically using \(\mathrm{Ce}^{4+}\) solution. When \(80 \% \mathrm{Fe}^{2+}\) is titrated, the EMF of the system in \(V\) is (Given, \(E^{\circ} \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.77 \mathrm{~V}\) and \(\left.\mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Ce}^{3+}\right)\) \((log 2=0.3, \log 3=0.5, \log 4=0.6)\)
- A 0.806
- B 0.532
- C 0.734
- D 0.756
Answer & Solution
Correct Answer
(C) 0.734
Step-by-step Solution
Detailed explanation
\(\because E=E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \frac{\left[\mathrm{Fe}^{3+}\right]}{\left[\mathrm{Fe}^{2+}\right]}\) Given, \(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.77 \mathrm{~V}\) and,…
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