In the following redox reaction, Cu ( OH )2(s)+ N 2 H 4(a q) arrow Cu (s)+ N 2( g ) Number of moles of Cu ( OH )2 reduced by 1 mole of N 2 H 4 is

Question

In the following redox reaction, Cu ( OH )2(s)+ N 2 H 4(a q) arrow Cu (s)+ N 2( g ) Number of moles of Cu ( OH )2 reduced by 1 mole of N 2 H 4 is (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/52e0d22fd022bde2437bca5f04a3c78c-.png" /> ∴ Thus, 1 mole N 2 H 4 ≡ 2 mol Cu 2+

Q. In the following redox reaction,
$C u (O H)_{2} (s) + N_{2} H_{4} (a q) \to C u (s) + N_{2} (g)$
Number of moles of $C u (O H)_{2}$ reduced by $1$ mole of $N_{2} H_{4}$ is

1

2

3

4

$∴$ Thus, $1$ mole $N_{2} H_{4} \equiv 2 m o l C u^{2 +}$

Q. In the following redox reaction, Cu(OH)2​(s)+N2​H4​(aq)→Cu(s)+N2​(g) Number of moles of Cu(OH)2​ reduced by 1 mole of N2​H4​ is

1

2

3

4

∴ Thus, 1 mole N2​H4​≡2molCu2+

Q. In the following redox reaction,
$C u (O H)_{2} (s) + N_{2} H_{4} (a q) \to C u (s) + N_{2} (g)$
Number of moles of $C u (O H)_{2}$ reduced by $1$ mole of $N_{2} H_{4}$ is

$∴$ Thus, $1$ mole $N_{2} H_{4} \equiv 2 m o l C u^{2 +}$