JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
For the given reactions \(Sn ^{2+}+2 e ^{-} \rightarrow Sn\) \(Sn ^{4+}+4 e ^{-} \rightarrow Sn\) The electrode potentials are; \(E _{ Sn ^{2+} / Sn }^{\circ}=-0.140\, V\) and \(E _{ Sn ^{4+} / Sn }^{\circ}=0.010\, V\). The magnitude of standard electrode potential for \(Sn ^{4+} / Sn ^{2+}\) i.e. \(E _{ Sn ^{4+} / Sn ^{2+}}^{\circ}\) is \(.....\times 10^{-2}\, V\). (Nearest integer)
- A \(320\)
- B \(32\)
- C \(16\)
- D \(160\)
Answer & Solution
Correct Answer
(C) \(16\)
Step-by-step Solution
Detailed explanation
\(Sn ^{2+}+2 e ^{-} \rightarrow Sn \quad \Delta G _{1}^{0}=+2 \times 0.140 \times F\) \(Sn ^{+4}+4 e ^{-} \rightarrow Sn \quad \Delta G _{2}^{0}=-4 \times 0.01 \times F\) ____________________________________________…
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