STATEMENT-1 : Magnesium can be prepared by the electrolysis of aq. $M g C l_{2}$
STATEMENT-2 : The reduction potential of $M g^{2 +}$ is much lower than that of $H^{+}$ .

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

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STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

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STATEMENT-1 is True, STATEMENT-2 is False

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STATEMENT-1 is False, STATEMENT-2 is True

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Solution

The correct option is D STATEMENT-1 is False, STATEMENT-2 is True
Magnesium can not be prepared by the electrolysis of aq. $M g C l_{2}$
as reduction potential of $M g^{2 +}$ is much lower than that of $H^{+}$ so on electrolysis of aq. $M g C l_{2},$ $H_{2}$ gas will be produced instead of $M g$ .

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Similar questions

- 1 :

0 < x < y \Rightarrow {log}_{a} x > {log}_{a} y

a > 1

- 2 :

a > 1, {log}_{a} x

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

n \geq 2,

\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} > \sqrt{n}

\sqrt{n (n + 1)} < n + 1

N H_{3}

P H_{3}

N H_{3}

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