Shorthand notation anode cathode1/22/2024 ![]() ![]() ![]() When the circuit is closed, electrons flow from the anode to the cathode. The oxidation half-reaction occurs at one electrode (the anode), and the reduction half-reaction occurs at the other (the cathode). (This is a repost because I was told that the TAs only look at the last four posts so I'm making sure I get my extra credit)\)). (To my TA: this is part 1/4 of my extended answers for extra credit) Divide 300,000 by 6,000 to obtain a value of 50. Let’s plug in 300,000 J for G o to match R. Even though G o is normally expressed as kJ/mol, R is expressed as J/molK, so we can convert R or G o to match units. ![]() That will help you place the electrons on the right side of the half-reaction. We can plug in the value of G o on the left side of the equation. Remember that the electrons flow from the anode to the cathode. Also, oxidation loses electrons, and reduction gains electrons. The best thing to remember is that oxidization takes place at the anode and reduction at the cathode. The two half-reactions have the same number of moles of electrons so that's good. The completed and balanced reduction half-reaction will be 2AgCl(s) + 2e-> 2Ag(s) + 2Cl. Because the cathode grows in mass, Ag(s) (the cathode) will be on the right side of the reduction half-reaction and AgCl(s) will be on the left side of the equation. In that case, Ag(s) will gain electrons and the Ag in AgCl(s) will be converted into Ag(s) because Ag's oxidation number goes from +1 to 0, indicating a gain in electrons. Electrons are pulled by the current to the cathode. Reduction gains electrons and so you know the electrons will be on the left side of the equation. Second, you would write the reduction half-reaction. We know that the electrodes should be at the far ends, hence. In this question, we should see the order as follows: Pb -> PbCl2 on the anodes surface -> Cl- aqueous in the anode solution -> Cl- aqueous in the cathode solution -> AgCl on the cathodes surface -> Ag. The completed and balanced oxidization half-reaction will be Pb(s) + 2Cl-> PbCl2(s) + 2e- In shorthand notation, the order goes from anode -> bridge -> cathode. Because the anode depletes, Pb(s) (the anode) will be on the left side of the oxidation half-reaction and PbCl2(s) (the oxidized state) will be on the right side of the equation. Spectator ions are not included in the notation. The concentrations of important species in each of the half reactions are included. The notation shows the anode on the left and cathode on the right. There is a shorthand notation used to specify the conditions of an electrochemical cell. Pb's oxidation number goes from 0 to +2, indicating a loss in electrons. Shorthand Notation for an Electrochemical Cell. In that case, Pb(s) will lose electrons and be in its oxidized state, PbCl2(s). Electrons are pulled away by the current to the cathode. Oxidation loses electrons and so you know the electrons will be on the right side of the equation. The right electrode (Ag) is the cathode and reduction takes place on that electrode.įirst, you would write the oxidation half-reaction. Remember the anode is on the left electrode (Pb) and oxidation takes place on that electrode. Then, think of the cathode half-reaction as a reduction to split the solid found on the cathode to its respective elements. As a general rule, think of the anode half-reaction as an oxidation to form the solid found on the anode. Since you are writing the reduction of AgCl in this case, you would see AgCl combining with the electrons to then split into 2Ag(s) and 2Cl. This means that for the half-reaction, you would see the combination of Pb(s) with the chlorine in solution to form the substance found on the anode which, in this case is PbCl2(s).įor the cathode half-reaction, the cathode is Ag(s). The electrode on the left is the anode, and the one on the right is the cathode.Ĭan someone explain how you write the anode and cathode half reaction?Screenshot at 8.07.35įor the anode, you see that there is Cl- in the solution and the anode is made of Pb(s). Maryam Alturki wrote:hello this question reads:Ĭomplete the half‑reactions for the cell shown, and show the shorthand notation for the cell.
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