1. Tardigrade
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  3. Chemistry
  4. The equilibrium constant for the reaction Zn ( s )+ Sn 2+( aq ) leftharpoons Zn 2+( aq )+ Sn ( s ) is 1 × 1020 at 298 K. The magnitude of standard electrode potential of Sn / Sn 2+ if E Zn 2+ / Zn 0=-0.76 V is × 10-2 V. (Nearest integer) Given: (2.303 RT / F )=0.059 V

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Q. 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 $E _{ Zn ^{2+} / Zn }^0=-0.76 \,V$ is ___$\times 10^{-2} \, V$. (Nearest integer)
Given: $\frac{2.303 \,RT }{ F }=0.059 \,V$

JEE MainJEE Main 2023Electrochemistry

Solution:

$ Zn ( s )+ Sn ^{2+}( aq ) \rightleftharpoons Zn ^{2+}( aq )+ Sn ( s ) $
$ \Delta G ^{\circ}=-2.303 RT \log _{10} Keq $
$ - nF \left( E _\text{ 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 $
$ 0.76+ E _{ Sn ^{2+} / Sn }^0=\frac{0.059}{2} \log _{10} 10^{20} $
$ 0.76+ E _{ Sn ^{2+} / Sn }^0=\frac{0.059 \times 20}{2} $
$ E _{ Sn ^{2+} / Sn ^0}=0.59-0.76=-0.17 $
$ E _{ Sn / Sn ^{2+}}^0=17 \times 10^{-2} V $
$ =17$