Two moles of electrons can do which of the following:
I. Reduce one mole of $A g^{+}$ to $A g$
II. Reduce one mole of $C u^{2 +}$ to $C u$
III. Reduce $\frac{2}{3}$ mole of $A l^{3 +}$ to $A l$

Question

Two moles of electrons can do which of the following:
I. Reduce one mole of

A g^{+}

to

A g

II. Reduce one mole of

C u^{2 +}

to

C u

III. Reduce

\frac{2}{3}

mole of

A l^{3 +}

to

A l

Toppr Admin · Accepted Answer

Cu2-2e-x2192-Cu1 mole of Cu2- requires 2 moles of electrons to reduce to Cu23Al3-2e-x2192-23Al23 moles of Al3- requires 2 moles of electrons to reduce to Al-Hence- option C is correct-

Two moles of electrons can do which of the following:
I. Reduce one mole of $A g^{+}$ to $A g$
II. Reduce one mole of $C u^{2 +}$ to $C u$
III. Reduce $\frac{2}{3}$ mole of $A l^{3 +}$ to $A l$
$I, I I, a n d I I I$
$I a n d I I o n l y$
$I o n l y$
$I I a n d I I I o n l y$

$C u^{2 +} + 2 e \to C u$

1 mole of $C u^{2 +}$ requires 2 moles of electrons to reduce to $C u$

$\frac{2}{3} A l^{3 +} + 2 e \to \frac{2}{3} A l$

$\frac{2}{3}$ moles of $A l^{3 +}$ requires 2 moles of electrons to reduce to $A l$ .

Hence, option $C$ is correct.

Two moles of electrons can do which of the following:I. Reduce one mole of Ag+ to AgII. Reduce one mole of Cu2+ to CuIII. Reduce 23 mole of Al3+ to AlI,II, and IIII and II onlyI onlyII and III only

Cu2++2e→Cu1 mole of Cu2+ requires 2 moles of electrons to reduce to Cu23Al3++2e→23Al23 moles of Al3+ requires 2 moles of electrons to reduce to Al.Hence, option C is correct.

Two moles of electrons can do which of the following:
I. Reduce one mole of $A g^{+}$ to $A g$
II. Reduce one mole of $C u^{2 +}$ to $C u$
III. Reduce $\frac{2}{3}$ mole of $A l^{3 +}$ to $A l$
$I, I I, a n d I I I$
$I a n d I I o n l y$
$I o n l y$
$I I a n d I I I o n l y$

$C u^{2 +} + 2 e \to C u$

1 mole of $C u^{2 +}$ requires 2 moles of electrons to reduce to $C u$

$\frac{2}{3} A l^{3 +} + 2 e \to \frac{2}{3} A l$

$\frac{2}{3}$ moles of $A l^{3 +}$ requires 2 moles of electrons to reduce to $A l$ .

Hence, option $C$ is correct.