The question is incomplete, the complete question is;
complete the half reactions for the cell shown here and show the shorthand notation for the cell by dragging labels to the correct position. The electrode on the left is the anode, and the one on the right is the cathode.
in the left side there is a solution of Pbcl2(s) and it contains the elctrode pb and on the right side there is solution of AgCl(S) which contains the electrode Ag. There is inverted glass tube with Kcl (aq) on the left and and Kcl(aq) on the right.
Anode half-reaction:
Blank + 2Cl^- equilibrium arrow yields Blank + 2e^-
Cathode half reaction:
Blank +2e^- equlibriums arrow yields Blank + 2Cl^-
Shorthand notation: Please provide.
Answer:
Explanation:
Let us remember that oxidation occurs at the anode while reduction occurs at the cathode. This will guide our work here. The Pb(s)/Pb^2+(aq) is the anode while the Ag^+(aq)/Ag(s) is the cathode
Anode half-reaction:
Pb(s) + 2Cl^- ⇄ Pb^2+(aq) + 2e^-
Cathode half reaction:
2Ag^+(aq) + 2e^- ⇄ 2Ag(s) + 2Cl^-
Overall reaction equation;
Pb(s) + 2Ag^+(aq) ⇄ Pb^2+(aq) + 2Ag(s)
Shorthand notation;
Pb(s)/Pb^2+(aq)//Ag^+(aq)/Ag(s)
