Match column I with column II

[Atomic masses : Cu= 63.5, Ag= 108, Zn= 65.4, Al=27]

A-r, B-p, C-s, D-q

The correct answer is option 2. A-r, B-p, C-s, D-q.

We know,

\(\text{Electrochemical equivalence (E.C.E.) = }\frac{\text{Equivalent weight}}{96500}\) ------ (i)

So, to find the electrochemical equivalence (E.C.E.), we need to find the equivalent weight of each element using the formula

\(\text{Equivalent Weight = }\frac{\text{Atomic mass of the element}}{\text{valency}}\)

A. \(Cu^{2+} (aq) +  2e^- \rightarrow Cu\)

\(∴ \text{Equivalent Weight = }\frac{63.5}{2} = 31.75\)

\(∴ \text{Electrochemical equivalence (E.C.E.) = }\frac{31.75}{96500} = 0.000329\)

Which corresponds to the value given in (r).

B. \(Ag^+ (aq) +  e^- \rightarrow Ag\)

\(∴ \text{Equivalent Weight = }\frac{108}{1} = 108\)

\(∴ \text{Electrochemical equivalence (E.C.E.) = }\frac{108}{96500} = 0.00112\)

Which corresponds to the value given in (p).

C. \(Zn^{2+} (aq) +  2e^- \rightarrow Zn\)

\(∴ \text{Equivalent Weight = }\frac{65.4}{2} = 32.75\)

\(∴ \text{Electrochemical equivalence (E.C.E.) = }\frac{32.75}{96500} = 0.000339\)

Which corresponds to the value given in (s).

D. \(Al^{3+} (aq) +  3e^- \rightarrow Al\)

\(∴ \text{Equivalent Weight = }\frac{27}{3} = 9\)

\(∴ \text{Electrochemical equivalence (E.C.E.) = }\frac{9}{96500} =0.000093\)

Which corresponds to the value given in (q).

So, the correct answer is option (2) A-r, B-p, C-s, D-q.