$\begin{array}{1 1}(a)\;1A&(b)\;1.5A\\(c)\;2A&(d)\;0.5A\end{array}$

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Let current of 1 ampere is passed through the solution for 1 hours.

$\therefore$ charge passed=$1\times 60\times 60=3600$

Mole of $e^-$ passed =$\large\frac{3600\times I}{96500}$

At anode :$2ON^-\rightarrow \large\frac{1}{2}$$O_2+H_2O+2e^-$

At cathode :$2H^++2e^-\rightarrow H_2$

Moles of $O_2$ released at anode $=\large\frac{3600\times I}{96500}\times \frac{1}{4}$

Volume of $O_2$ releases =$\large\frac{3600\times I}{96500}\times \frac{1}{4}$$\times 22400$

$\Rightarrow 208.9I$

Moles of $H_2$ released at cathode $=\large\frac{3600\times I}{96500}\times \frac{1}{2}$

Volume of $H_2$ releases =$\large\frac{3600\times I}{96500}\times \frac{1}{2}$$\times 22400$

$\Rightarrow 417.8I$

Total volume of gas released at S.T.P =208.9I+417.8I=338ml

I=0.5amp

Hence (d) is the correct answer.

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