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(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K :

$Sn (s) | Sn^{2+} (0.004 M) || H^{ +} (0.020 M) | H_{2} (g) (1 bar) | Pt (s)$

(Given : $E^{\circ}_{Sn^{2+}/Sn}=-0.14V$ )

(b) Give reasons :

(i) On the basis of $E^{\circ}$ values, $O_{2}$ gas should be liberated at anode but it is $Cl_{2}$ gas which is liberated in the electrolysis of aqueous NaCl.

(ii) Conductivity of $CH_3COOH$ decreases on dilution.

Answers (1)

(a)

$Sn (s) | Sn^{2+} (0.004 M) || H^{ +} (0.020 M) | H_{2} (g) (1 bar) | Pt (s)$

$E^{\circ}_{Sn^{2+}/Sn}=-0.14V$

$Sn(s)\rightarrow Sn^{2+}(aq)+2e^{-}$

$2H^{+}+2e^{-}\rightarrow H_{2}$

__________________________________________

$Sn(s)+2H^{+}(aq)\rightarrow Sn^{2+}(aq)+H_{2}(g)$

___________________________________________

$E^{\circ}_{cell}=0-(-0.14V)$

$E^{\circ}_{cell}=0.14V$

$E_{cell}=E^{\circ}_{cell}-\frac{0.0591}{n}\log \frac{[Sn^{2+}]}{[H^{+}]^{2}}$

$E_{cell}=0.14-\frac{0.0591}{2}\log \frac{(4\times 10^{3-})}{(2\times 10^{2})^{2}}$

$E_{cell}=0.14-\frac{0.0591}{2}\log 10$

$E_{cell}=0.14-0.0295$

$E_{cell}=0.1105V$

(b)

(i) On the basis of $E^{\circ}$ values, $O_{2}$ gas should be liberated at anode but it is $Cl_{2}$ gas which is liberated in the electrolysis of aqueous NaCl. This is because higher voltage is required to liberate oxygen gas which is greater than calculated from standard potential.

(ii) Conductivity of $CH_3COOH$ decreases on dilution since with dilution, concentartion decreases with decreases the number of ions / volume.

Posted by

Sumit Saini

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