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  • For an electrochemical cell <img alt="\mathrm{Sn(s)\left | Sn^{2+}(aq,1M) \right |\left | Pb^{2+}(aq,1M) \right |Pb(s)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BSn%28s%29%5Cleft%20%7C%20Sn%5E%7B2&plus;%7D%28aq%2C1M%29%20%5C

For an electrochemical cell \mathrm{Sn(s)\left | Sn^{2+}(aq,1M) \right |\left | Pb^{2+}(aq,1M) \right |Pb(s)} the ratio \mathrm{\frac{\left [ Sn^{2+} \right ]}{\left [ Pb^{2+} \right ]}} When this cell attains equilibrium is _____. \mathrm{( Given :E^{0}_{Sn^{2+} | Sn} =-0.14V,} \mathrm{E^{0}_{Pb^{2+} | Pb }=-0.13V,} \mathrm{\frac{2.303RT}{F}=0.06)}
Option: 1 2.1544
Option: 2 1.11
Option: 3 7.15
Option: 4 3.14
 

Answers (1)

best_answer

As we have learnt, 

Nernst equation is given as 

\mathrm{E= E^0_{cell}-\frac{2.303RT}{nF}logQ}

Now, the chemical reaction occuring in the cell is given as 

\begin{array}{l}{\mathrm{Sn}+\mathrm{Pb}^{2+} \longrightarrow \mathrm{Sn}^{2+}+\mathrm{Pb}} \\ {0=0.01-\frac{0.06}{2} \log \left\{\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\right\}} \\ {0.01=\frac{0.06}{2} \log \left[\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\right\}} \\ {\frac{1}{3}=\log \left[\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\right\} \Rightarrow \frac{\left[\mathrm{Sb}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}=10^{1 / 3}=2.1544}\end{array}

Hence, the option number (1) is correct.

Posted by

Kuldeep Maurya

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