The quantity of electricity required to reduce 12.3 g of nitro benzene to aniline assuming 50% current efficiency is:

115800 C

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57900 C

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231600 C

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28950 C

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Solution

The correct option is A 115800 C
Reaction for reduction :

$C_{6} H_{5} {N O}_{2} + {6 H}^{+} + 6 e^{-} ⟶ C_{6} H_{5} {N H}_{2} + 2 H_{2} O$

Given, molar mass of $C_{6} H_{5} {N O}_{2} = 123$

Here, $12.3 g$ nitrobenzene mole $= \frac{6 \times 12.3}{123} = 0.6$

Charge $= 0.6 \times 96500$ coloumbs
$= 57900$

Efficiency $= 50$ %

$\Rightarrow$ $2 \times 57900 =$ coloumbs

$∴$ charge required is $115800$ coloumbs.

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The quantity of electricity required to reduce 12.3 g of nitro benzene to aniline assuming 50% current efficiency is:

The correct option is A 115800 CReaction for reduction :C6H5NO2+6H++6e−⟶C6H5NH2+2H2OGiven, molar mass of C6H5NO2=123Here, 12.3 g nitrobenzene mole =6×12.3123=0.6Charge =0.6×96500 coloumbs =57900Efficiency =50%⇒ 2×57900= coloumbs∴ charge required is 115800 coloumbs.