Question

The oxide of a metal (R) can be reduced by the metal (P) and metal (R) can reduce the oxide of metal (Q). Then the decreasing order of the reactivity of metal (P), (Q) and (R) with oxygen is:
A. $P$$ >$$ Q$$ > $$R$
B. $P $$>$$ R$$ > $$Q$
C. $R $$> $$P$$ > $$Q$
D. $Q$$ >$$ P $$> $$R$

Answer 1

Hint: The element which can reduce the oxides of the other element must be more reactive than the other element. Hence, the element which is more reactive will replace the other element.

Complete step by step solution:
If you look at first statement then it says that, oxide of metal R can be reduced by metal P therefore,$RO+P\to PO+R$.........................(1)
Since, this equation is happening or spontaneous, which means its $\Delta {{G}_{1}}<0$
If we take other equations such as:
$P+O\to PO$......................(2)
$RO\to R+O$.......................(3)
$R+O\to RO$.......................(4)
Then according to the equation above, if we would add the equation 2 and 3 then we would get equation 1,
$\Rightarrow \Delta {{G}_{2}}+\Delta {{G}_{3}}<0$
But, if you look closely then you will notice that equation 4 Is just the opposite of equation 3 therefore, we can replace it,
$\Rightarrow \Delta {{G}_{2}}-\Delta {{G}_{4}}<0$ $(\Delta {{G}_{3}}=-\Delta {{G}_{4}})$
$\Rightarrow \Delta {{G}_{2}}<\Delta {{G}_{4}}$
The above conclusion shows that reaction 2 is much more stable than 4, hence, PO will be formed more which means that P is reactive than R. We can do the same with the other statement where the product will be RO which will conclude that R is much more reactive than Q .

Therefore, the answer of this question is B. $P$$ >$$ R$$ >$$ Q$.

Note: You can also understand this question with the help of reactivity series of a metal where the metals who are at higher place can only displace the other metal. In this question when P reduces the oxide, it means it must be at a higher place in reactivity series than R and Q.