You visited us 1 times! Enjoying our articles? Unlock Full Access!

Oxidising Reagent and Reducing Reagent

These equilib...

Question

These equilibria can be discussed in terms of thermodynamic functions.
For the reaction :

(i) $M_{(s)} + \frac{1}{2} O_{2 (g)} \to M O_{(} s)$
(ii) $\frac{1}{2} C_{(s)} + \frac{1}{2} O_{2 (g)} \to \frac{1}{2} C O_{2 (g)}$
(iii) $C_{(s)} + \frac{1}{2} O_{2 (g)} \to C O_{(g)}$
(iii) $C O_{(g)} + \frac{1}{2} O_{2 (g)} \to C O_{2 (g)}$

The temperature dependence of $Δ ˚ G$ of reaction (i) to (iv) is shown in the given diagram. This is known as the Ellingham diagram. With the help of the Ellingham diagram, one can easily predict the most suitable reducing agent for the reduction of metal oxides.

$A I_{2} O_{3}$ can be reduced by carbon at a temperature of:

30^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

500^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2000^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

> 2500^{\circ}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $> 2500^{\circ}$ C
The Ellingham diagram for oxides shows several important features:
(a)The graph for metal oxide all slope upwards, because the free energy change increases with an increase of temperature as discussed above.
(b)The free energy changes all follows a straight line unless the materials metal or vaporize.
(c)When the temperature is raised, a point will be reached where the graph crosses the $Δ G = 0$ line.
Below this temperature the free energy of formation of the oxide is negative, so the oxide is stable. Above this temperature the free of formation of the oxide is positive, and the oxide becomes unstable, and should decompose into the metal and dioxygen.
(d)Any metal will reduce the oxide of other metals which lie above it in the Ellingham diagram because the free energy will become more negative by an amount equal to the different between the two graphs at that particular temperature.
(e) Al metal can reduce the oxide of metal placed above in the diagram. Due to more negative free energy Al can reduce Zn but not Mg.
From figure it can be seen that,
Alumina can be reduced by carbon at a temperature > 2500 C.

Suggest Corrections

Similar questions

C (s) + \frac{1}{2} O_{2} (g) \to C O (g); Δ H_{1} = a

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g); Δ H_{2} = b

C (s) + C O_{2} (g) \to 2 C O (g); Δ H_{3} = c

^{\circ} C

Z n_{(s)} + 1 / 2 O_{2 (g)} ⟶ Z n O_{(s)}, Δ G^{\circ} = - 360 k J m o l^{- 1}

C_{(s)} + 1 / 2 O_{2 (g)} ⟶ C O_{(g)}, Δ G^{\circ} = - 460 k J m o l^{- 1}

^{\circ} C

Δ H_{c o m b} C

I

C (s) + O_{2} (g) ⟶ C O_{2} (g)

I I

C (s) + \frac{1}{2} O_{2} (g) ⟶ C O (g)

I I I

C O (g) + \frac{1}{2} O_{2} (g) ⟶ C O_{2} (g)

M O_{(s)}

C

+ O_{2} (g) \to C O_{2} (g)

Δ_{r} H^{o} = x k J m o l^{- 1}

C

+ \frac{1}{2} O_{2} (g) \to C O (g)

Δ_{r} H^{o} = y k J m o l^{- 1}

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)

Δ_{r} H^{o} = z k J m o l^{- 1}

Join BYJU'S Learning Program