A Write the cell reaction and calculate the e-m-f of the following cell at 298 K -Sns Sn 2-0-004 M H -0-020 M - H 2g1 bar - Pt s Given - E Sn 2- - Sn -0-14 V b Give reasons-i On the basis of E - 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 3 COOH decreases on dilution-

(a) E^∘_cell = nbsp;E^∘_red nbsp;E^∘_oxi = (0) ( 0.14) = 0.14 V E_cell = E^∘_cell 0.0591/n log [Sn^2+]/[H+]^2 = 0.14 0.0591/2 nbsp; log nbsp ...

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Electrometallurgy

a Write the c...

Question

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

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

(Given : $E_{S n^{2 +} / S n}^{\circ}$ = -0.14 V)

(b) Give reasons :

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

(ii) Conductivity of $C H_{3} C O O H$ decreases on dilution.

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S n (s) | S n^{2 +} (0.004 M) | | H^{+} (0.020 M) | H_{2} (g) (1 b a r) | P t (s)

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