Calculate the quantity of electricity that would be required to reduce 12.3 g of nitrobenzene to aniline, if the current efficiency for the process is 50 percent. If the potential drop across the cell is 3.0 volts, how much energy will be consumed?

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Solution

The reduction reaction is: $C_{6} H_{5} N O_{2} + 6 H^{+} + 6 e^{-} \to C_{6} H_{5} N H_{2} + 2 H_{2} O$
Mass of $C_{6} H_{5} N O_{2}$ reduced is $12.3 g$
Amount of $C_{6} H_{5} N O_{2}$ to be reduced $= \frac{12.3}{M (C_{6} H_{5} N O_{2})} = \frac{12.3}{123} = 0.1 m o l$
To reduced $0.1 m o l$ of $C_{6} H_{5} N O_{2}$ $0.6 m o l$ of electrons are required.
But current efficiency is only $50 %$ .
Hence the amount of electron required would be $1.2 m o l$ .
$∴$ Electricity carried by $1.2 m o l$ of electrons $= 1.2 \times 96500 = 115800 C$
Energy consumed $= 115800 \times 3.0 = 347400 J = 347.4 K J$

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