JEE MAIN - Chemistry (2023 - 29th January Evening Shift - No. 17)
The equilibrium constant for the reaction
$$\mathrm{Zn(s)+Sn^{2+}(aq)}$$ $$\rightleftharpoons$$ $$\mathrm{Zn^{2+}(aq)+Sn(s)}$$ is $$1\times10^{20}$$ at 298 K. The magnitude of standard electrode potential of $$\mathrm{Sn/Sn^{2+}}$$ if $$\mathrm{E_{Z{n^{2 + }}/Zn}^\Theta = - 0.76~V}$$ is __________ $$\times 10^{-2}$$ V. (Nearest integer)
Given : $$\mathrm{\frac{2.303RT}{F}=0.059~V}$$
Answer
17
Explanation
$$E_{cell}^o = {{2.303\,RT} \over {2F}}\log k$$
$$E_{cell}^o = {{0.059} \over 2}\log ({10^{20}})$$
$$E_{Z{n^{2 + }}/Zn}^o + 0.76 = 0.59$$
$$E_{Z{n^{2 + }}/Zn}^o = 0.59 - 0.76$$
$$E_{Zn/Z{n^{2 + }}}^o = 0.17\,V$$
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