Each of the given beakers contains two metal strips of same size fastened together and immersed in in hydrochloric acid.

After 5 minutes, amount of Q ions formed is greater than P ions in beaker 1, S ions is greater than R ions in beaker 2, R ions is greater than Q ions in beaker 3, and T ions is greater than S ions in beaker 4.
Correct order of reactiivity of these metals is

Options

A. `P gt Q gt R gt S gt T`
B. `R lt P lt T lt S lt Q`
C. `T gt R gt Q gt P gt S`
D. `P lt Q lt R lt S lt T`

Answer

D

Explanation

In beaker 1, Q ions > P ions
`therefore` Reactivity of Q>P with hydrochloric acid.
In beaker 2, S ions > R ions
`therefore`Reactivity of S>R with hydrochloric acid.
In beaker 3, R ions >Q ions
`therefore` Reactivity of R>Q with hydrochloric acid.
In beaker 4, T ions >S ions
`therefore` Reactivity of `T gt S` with hydrochloric acid.
Thus, the correct order of reactivity of these metals is `P lt Q lt R lt S lt T`.