Question

Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at stp (1atm

Answer 1

2 CH2 COOK + 2H2O  -------> C2H4 (anode) + 2CO2 + H2 + 2KOH (cathode)

Equivalents of products (C2H4 + CO2 + H2) = 0.2 + 0.2 + 0.2 = 0.6

n of gases = 0.2/2 + 0.2/1 + 0.2/2 = 0.4

As, PV = nRT

Then,

V = nRT/P

V = 0.4 x 0.0821 x 273 / 1

V = 8.96 L

Answer 2

Hey dear,

◆ Answer-
V = 8.964 L

◆ Explaination-
Oxidation of succinate ion occurs as follows-
CH2COO- + H2O  ---> 1/2 C2H4 + CO2 + 1/2 H2 + OH-

According to Faraday's law in electrochemistry, mass equivalents fir each gas is 0.2 .

Actual no of moles is calculated by-
n = 0.2×1/2 + 0.2×1 + 0.2×1/2
n = 0.4

At STP 1 mol of gas is 22.41 L
Thus, 0.4 mol => 0.4×22.41 = 8.964 L.

Therefore, total volume of gases produced is 8.964 L.

Hope it helps...