$\begin{array}{ll}\text{LIST - 1}& \text{ LIST - 2 }\\ \text{A) One faraday}& \text{1) Reduction}\\ \text{B) Anode}& \text{2) 96500 coulomb}\\ \text{C) Cathode}& \text{3)}6.24\times {10}^{18}\text{ electrons}\\ \text{D) One coulomb}& \text{4) Oxidation}\\ & \text{5) }\mathrm{Z}\times \text{96,500}\end{array}$
The correct match is

Faraday is defined as the charge carried by one mole of electrons = 96500 coulombs.

Because oxidation entails the loss of electrons at the anode, oxidation happens.

Because reduction includes electron obtaining, electrons are always acquired at the cathode.