$2 N a (s) + {C l}_{2} (g) ⟶ 2 N a C l (s)$
1

Open in App

Solution

The correct option is A 1
The reaction is given below:
$2 N a (s) + {C l}_{2} (g) ⟶ 2 N a C l (s)$
0 0 +1 -1
$N a$ is oxidised to ${N a}^{+}$ and therefore, ${C l}_{2}$ is the oxidising agent (oxidiser).
${C l}_{2}$ is reduced to ${C l}^{⊖}$ and therefore, $N a$ is the reducing agent (reducer).

Suggest Corrections

Similar questions

2 N a (s) + {C l}_{2} (g) ⟶ 2 N a C l (s)

2 N a (s) + {C l}_{2} (g) ⟶ 2 N a C l (s)

H_{2} (g) + C l_{2} (g) \to 2 H C l (g)

Δ H = - 44 K c a l

2 N a (s) + 2 H C l (g) \to 2 N a C l (s) + H_{2} (g)

Δ H = - 152 K c a l

N a (s) + \frac{1}{2} C l_{2} (g) \to N a C l (s)

Δ H = ?

H_{2} (g) + C l_{2} (g) \to 2 H C l (g); Δ H^{o} = - 44 K c a l

2 N a (s) + 2 H C l (g) \to 2 N a C l (s) + H_{2} (g);

Δ H = - 152 K c a l

N a (s) + 0.5 C l_{2} (g) \to N a C l (s); Δ H^{o} = ?

2 N a (s) + C l_{2} (g) \to 2 N a C l (s) + 822 k J

0.5 m o l

Join BYJU'S Learning Program