Electrolysis of an acetate solution produces ethane according to the Kolbe reaction: 2CH3COO- arrow C2H6+2CO2+2e- What volume of ethane is produced at 27° C and 740 mm Hg , if a current of 0.5 ampere were passed through the solution for 7 hours and the electrode reaction is 82 % efficient?

Question

Electrolysis of an acetate solution produces ethane according to the Kolbe reaction: 2CH3COO- arrow C2H6+2CO2+2e- What volume of ethane is produced at 27° C and 740 mm Hg , if a current of 0.5 ampere were passed through the solution for 7 hours and the electrode reaction is 82 % efficient? (A) 2.70 L (B) 5.4 L (C) 1.35 L (D) 0.65 L

Tardigrade Admin · Accepted Answer

Number of moles of e-=(0.500 × 7.00 × 3600/96500)× 0.82=0.1071 mole Two moles of electrons are associated with 1 mole C2H6, thus, Volume of C2H6=(0.1071/2)× 22.4× (760/740)× (300/273)=1.35 L

$2.70$ $L$

$5.4$ $L$

$1.35$ $L$

$0.65$ $L$

Number of moles of $e^{-} = \frac{0.500 \times 7.00 \times 3600}{96500} \times 0.82 = 0.1071$ $m o l e$
Two moles of electrons are associated with $1$ $m o l e C_{2} H_{6},$ thus,
Volume of $C_{2} H_{6} = \frac{0.1071}{2} \times 22.4 \times \frac{760}{740} \times \frac{300}{273} = 1.35$ $L$

2.70 L

5.4 L

1.35 L

0.65 L

Number of moles of e−=965000.500×7.00×3600​×0.82=0.1071 mole Two moles of electrons are associated with 1 moleC2​H6​, thus, Volume of C2​H6​=20.1071​×22.4×740760​×273300​=1.35 L

$2.70$ $L$

$5.4$ $L$

$1.35$ $L$

$0.65$ $L$

Number of moles of $e^{-} = \frac{0.500 \times 7.00 \times 3600}{96500} \times 0.82 = 0.1071$ $m o l e$
Two moles of electrons are associated with $1$ $m o l e C_{2} H_{6},$ thus,
Volume of $C_{2} H_{6} = \frac{0.1071}{2} \times 22.4 \times \frac{760}{740} \times \frac{300}{273} = 1.35$ $L$