Cd(s)/Cd(NO3)^2(aq) // AgNO^3 (aq)/ Ag(s)

Question

Electrodes, electroyltes and flow of electrons, ion movement

Answer 1

In the first reaction, Cd(s) (a solid electrode) is oxidized to Cd2+ ions in solution (electrolyte) and loses electrons. The Cd2+ ions then move through the electrolyte towards the cathode (negative electrode).

Cd(s) → Cd2+(aq) + 2e-

In the second reaction, Ag+ ions from AgNO3 solution (electrolyte) are reduced at the cathode (negative electrode) to Ag atoms, gaining electrons. The Ag atoms then deposit on the cathode (metallic Ag).

Ag+(aq) + e- → Ag(s)

Overall, the Cd(s) electrode is oxidized and provides electrons to the circuit, while the Ag+ ions are reduced and accept those electrons to form metallic Ag at the cathode.