Question

Use the following data to place the fictitious metals Q,R,T,X,A, and E into an activity series. Show your work.

R + A⁺ -->N.R.

Q+T⁺ --> T + Q⁺

E+T⁺ --> N.R.

R + E⁺ --> N.R.

E+X⁺ --> E⁺ + X

E +Q⁺ --> N.R.

X⁺ + A --> A + X⁺

E + A⁺ --> E⁺+ A

Answer 1

Solution :-

The compound which capable of reducing the other species is considered as more active.

From the data of the redox equations it shows that

R<A, Q>T, E<T, R<E, E>X, E<Q, A<X, E>A

R is less reactive than A because it cannot reduce the A^+

Q is more reactive than T because Q reduces the T^+

E is less reactive than T because it cant reduce the T^+

R is less reactive than E because it cant reduce the E^+

E is more reactive than X because it reduces the X^+

E is less reactive than Q because it cant reduce the Q^+

A is less reactive than X because it cant reduce the X^+

E is more reactive than A because it reduces the A^+

R is the least reactive and Q is the most reactive

Hence the order of activity is R < A < X < E < T < Q