Question

Electrolytic reduction of 6.15g of nitrobenzene using current efficiency of 40% will require how much quantity of electricity. (C=12,H=1,N=14,O=16)

Answer 1

No. of equivalents of nitrobenzene to aniline change in oxidation no. of N
 is 6=wt. in g equiv.
 wt. = 6.15/ 123/6 = 0.3mole of electricity for 100% efficiency = 0.3 F
but for 40% efficiency = 0.75 F= 0.75 ×96500 coulombs= 72375 C
so the answer is 0.75 F or 72375 C. 

Answer 2

Answer : The quantity of electricity required is 0.75 F

Explanation :

The reduction reaction is written as:

$C_6H_5NO_2+6H^++6e^-\rightarrow C_6H_5NH_2+3H_2O$

First we have to calculate the equivalent weight of nitrobenzene.

$\text{Equivalent weight of nitrobenzene}=\frac{\text{Molar mass of nitrobenzene}}{n}$

where,

n = valency factor = 6

$\text{Equivalent weight of nitrobenzene}=\frac{123}{6}$

Now we have to calculate the number of equivalents of nitrobenzene to aniline change in oxidation number of N is, 6.

$\text{Number of equivalents of nitrobenzene}=\frac{\text{weight in grams}}{\text{Equivalent weight of nitrobenzene}}=\frac{6.15}{(\frac{123}{6})}$

Quantity of electricity in coulombs = Current × Time = $\frac{6.15}{(\frac{123}{6})}$

For 40 %, Quantity of electricity in coulombs = $(Current\times Time)\frac{100}{40}=\frac{6.15}{(\frac{123}{6})}\times \frac{100}{40}=0.75F$

Therefore, the quantity of electricity required is 0.75 F