Question

For an electrochemical cell $$Sn(s)|{Sn}^{2+} (aq.1M)\parallel {Pb}^{2+} (aq. 1M)|Pb(s)$$, the ratio $$\cfrac { \left[ { Sn }^{ 2+ } \right] }{ \left[ { Pb }^{ 2+ } \right] } $$ when this cell attains equilibrium is ________ .

[Given: $${ E }_{ { Sn }^{ 2+ }|Sn }^{ 0 }=-0.14V,{ E }_{ { Pb }^{ 2+ }|Pb }^{ 0 }=-0.13V,\ \cfrac { 2.303RT }{ F } =0.06$$]

Answer 1

Correct option is A. 2.15
Given $$E^0_{{Sn^{2+}}/Sn}=-0.14\ V$$

$$E^0_{{Pb^{2+}}/Pb} = \ -0.13\ V$$

For cell reaction $$Sn|Sn^{2+}||Pb^{2+}|Pb$$

$$\dfrac{2.303RT}{F}=0.06$$

At equilibrum state, $$E_{cell}=0,E^0_{cell}=-0.13-(-0.14)=0.01\ V$$

$$Sn+Pb^{2+}\rightarrow Sn^{2+}+Pb$$

$$E=E^0-\dfrac{0.06}{n}\log \{\dfrac{[Sn^{2+}]}{[Pb^{2+}]}\}$$

$$\Rightarrow 0=0.01-\dfrac{0.06}{2}\log \{\dfrac{[Sn^{2+}]}{[Pb^{2+}]}\}$$

$$\Rightarrow 0.01=\dfrac{0.06}{2}\log \{\dfrac{[Sn^{2+}]}{[Pb^{2+}]}\}$$

$$\Rightarrow \dfrac{0.01\times 2}{0.06}=\log \dfrac{[Sn^{2+}]}{[Pb^{2+}]}$$

or $$\dfrac{[Sn^{2+}]}{[Pb^{2+}]}=10^{1/3}=2.1544$$