JEE MAIN - Chemistry (2016 - 10th April Morning Slot - No. 12)
Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :
2.24 L
4.48 L
6.72 L
8.96 L
Explanation
The reaction is
$$2C{H_3}COOK + 2{H_2}O\buildrel {Electrolysis} \over \longrightarrow C{H_3} - C{H_3} + 2C{O_2} + {H_2} + 2KOH$$
At the anode (Oxidation):
_10th_April_Morning_Slot_en_12_1.png)
At the cathode (Reduction):
$$2{H_2}O\buildrel { + 2{e^ - }} \over \longrightarrow 2O{H^ - } + 2\mathop H\limits^ \bullet $$
$$2\mathop H\limits^ \bullet \buildrel {} \over \longrightarrow {H_2}$$
Total number of moles of gases
= moles of C2H6 + moles of CO2 + moles of H2
$$n = {{0.2} \over 2} + {{0.2} \over 1} + {{0.2} \over 2} = 0.4$$
$$V = {{nRT} \over p}$$
$$ = {{(0.4 \times 0.0821 \times 273)} \over 1} = 8.96$$ L
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