JEE MAIN - Chemistry (2018 - 15th April Morning Slot - No. 13)
When an electric currents passed through acidified water, 112 mL of hydrogen gas at N.T.P. was collected at the cathode in 965 seconds. The current passed, in ampere, is :
1.0
0.5
0.1
2.0
Explanation
Reaction at cathode :
2H+ + 2e$$-$$ $$ \to $$ H2
We know,
$$\omega $$ = zIt = $${{EIt} \over {96500}}$$
<
no. of moles of H2 = $${{112} \over {22400}}$$
$$\therefore\,\,\,$$ mass (w) of H2 = $${{112} \over {22400}}$$ $$ \times $$ 2
$$\therefore\,\,\,\,$$ $${{112} \over {22400}}$$ $$ \times $$ 2 = $${{1 \times I \times 965} \over {96500}}$$
$$ \Rightarrow $$$$\,\,\,\,$$ $$I$$ = 1 A
2H+ + 2e$$-$$ $$ \to $$ H2
We know,
$$\omega $$ = zIt = $${{EIt} \over {96500}}$$
<
no. of moles of H2 = $${{112} \over {22400}}$$
$$\therefore\,\,\,$$ mass (w) of H2 = $${{112} \over {22400}}$$ $$ \times $$ 2
$$\therefore\,\,\,\,$$ $${{112} \over {22400}}$$ $$ \times $$ 2 = $${{1 \times I \times 965} \over {96500}}$$
$$ \Rightarrow $$$$\,\,\,\,$$ $$I$$ = 1 A
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