CBSE Class 12 Chemistry 2018 Solved Paper

Question : 25

Total: 26

(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298K :
Sn (s) | Sn2+(0.004M)||H+(0.020M)∣H2 (g) (1 bar) | Pt (s)
(Given : E°Sn2+∕Sn=−0.14V )
(b) Give reasons:
(i) On the basis of E∘ values, O2 gas should be liberated at anode but it is Cl2 gas which is liberated in the electrolysis of aqueous NaCl.
(ii) Conductivity of CH3‌COOH decreases on dilution.
OR
(a) For the reaction
‌2‌AgCl( s)+H2(g)(1‌atm)→2‌Ag( s)+2H+(0.1M)+2Cl−(0.1M),
‌∆G∘=−43600J‌ at ‌25∘C.
‌‌ Calculate the e.m.f. of the cell. ‌
‌[log‌10−n=−n]
Calculate the e.m.f. of the cell.
[log‌10−n=−n]
(b) Define fuel cell and write its two advantages.

Solution:

Sn (s) | Sn2+(0.004M)||H+(0.020M)∣H2 (g) (1 bar) | Pt (s)

‌Sn( s)→Sn2+(aq)+2e−
2H+(aq)+2e−→H2(g)
────────────────────
Sn( s)+2H+(aq)→Sn2+(aq)+H2(g)
────────────────────
E°‌cell ‌=E°(H+∕H2)−E°(Sn2+∕Sn)
‌=0.00−(−0.14)
‌=+0.14V
E‌cell ‌‌=E°‌cell ‌−‌

0.0591

‌log‌

[Sn2+]

[H+]2

‌=0.14−‌

0.0591

‌log‌

(4×10−3)

(2×10−2)2

‌=0.14−0.0295‌log‌10
‌=0.14−0.0295
‌=0.1105V
(b) (i) ‌NaCl→Na++Cl−
‌H2O⇌H++OH−
The value of E∘ of O2 is higher than Cl2 but O2 is evolved from H2O only when the higher voltage is applied. So, because of this Cl2 is evolved instead of O2.
(ii) Conductivity varies with the change in the concentration of the electrolyte. The number of ions per unit volume decreases on dilution. So, conductivity decreases with decrease in concentration. Therefore, conductivity of CH3‌COOH decreases on dilution.
OR
‌‌ (a) ‌∆G∘=−nFE∘
‌−43600=−2×96500×E∘
‌E∘=0.226V
‌E=Eo−0.059∕2‌log ([H+]2[Cl−]2∕[H2])
‌=0.226−0.059∕2‌log[(0.1)2 ×(0.1)2]∕1
‌=0.226−0.059∕2‌log‌10−4
‌=0.226+0.118=0.344V
(b) Cells that convert the energy of combustion of fuels (like hydrogen, methane, methanol etc.) directly into electrical energy are called fuel cells.
Advantages : High efficiency, non polluting (or any other suitable advantage)

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