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JEE Mains · Chemistry · STD 12 - 2. Electrochemistry

The equilibrium constant for the reaction \(Zn ( s )+ Sn ^{2+}( aq ) \rightleftharpoons Zn ^{2+}( aq )+ Sn ( s )\) is \(1 \times 10^{20}\) at \(298 K\). The magnitude of standard electrode potential of \(Sn / Sn ^{2+}\) if \(E _{ Zn ^{2+/ / nn }}^0=-0.76 V\) is \(............\times 10^{-2}\,V\). (Nearest integer) Given : \(\frac{2.303 RT }{ F }=0.059\,V\)

  1. A \(16\)
  2. B \(15\)
  3. C \(17\)
  4. D \(14\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(17\)

Step-by-step Solution

Detailed explanation

\(Zn ( s )+ Sn ^{2+}( aq ) \rightleftharpoons Zn ^{2+}( aq )+ Sn ( s )\) \(\Delta G ^0=-2.303\,RT \log _{10} Keq\) \(- nF \left( E _{ cell }^0\right)=-2.303\,RT \log _{10} Keq\) \(E _{ Zn / Zn ^{2+}}^0+ E _{ Sn ^{2+} / Sn }^0=\frac{0.059}{2} \log _{10} Keq\)…
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