For the following question, the answer is an integer between 0 and 9 (both inclusive).
In the Ellingham diagram for oxides, there are x lines for the reaction(s) between C and O. What is x?
In the Ellingham diagram for oxides, the general trend for the graphs for metal to metal oxide is
There is no common trend
They slope upwards
They are all parallel to the ∆Gº = 0 horizontal axis
They slope downwards like the 2C + O2 ⟶ 2CO line
- Positive value of ΔfG∘ for MxO indicates, MxO decompose on its own
- Metal with most -ve value of ΔG∘ for formation of oxide can act as a better reducing agent
- ΔG∘ is plotted for the oxidation of metal with 1 mole of O2
- ΔG∘ is plotted for reduction of 1 mole of metal oxide
You are given an Ellingham diagram (deliberately not shown). It is given that for almost all practically achievable temperatures, the Fe→Fe2O3 line lies above the Al→Al2O3 line and ΔfG∘ for both Al2O3 and Fe2O3 are negative. Which of the following statements is/are true?
Under the given conditions, neither metal can reduce the other oxide
Both b and c
Fe can reduce Al2O3 into Al
Al can reduce Fe2O3 into Fe
- Tin from SnO2
- Iron from Fe2O3
- Aluminium from Al2O3
- Magnesium from MgCO3.CaCO3
The ΔfG∘T line for C+O2 → CO2 is nearly parallel to the ΔfG∘ = 0 – True or False?
Which of the following is incorrect on the basis of the above Ellingham diagram for carbon?
- Up to 710∘C, the reaction of formation of CO2 is energetically more favorable, but above 710∘C, the formation of CO is preferred.
- In principle, carbon can be used to reduce any metal oxide at a sufficiently high temperature.
- Carbon reduces many oxides at elevated temperature because ΔG∘ vs temperature line has a negative slope.
- Reduction with carbon
- Reduction with aluminium
- Electrolytic reduction
- Reduction with CO
- 3TiO2(s) +4Al(s) → 2Al2O3 (s) +3Ti(l)
The metal that cannot be extracted by reduction of its oxide with Aluminium is: