Given the following half-reactions, Pb(s) + 21^-(aq) rightarrow PbI_2(s) + 2e^-
ID: 1060180 • Letter: G
Question
Given the following half-reactions, Pb(s) + 21^-(aq) rightarrow PbI_2(s) + 2e^- I_2(aq) + 2e^- rightarrow 21^-(aq) Write the balanced overall reaction. a. Pb(s) + I_2(aq) + 2e^- rightarrow PbI(s) + I^-(aq) b. Pb(s) + I_2(aq) rightarrow PbI_2(s) c. Pb(s) + CI_2(aq) + rightarrow Pb^2+(aq) + 2I^-(aq) d. 2Pb(s) + 4e^- rightarrow 2PbI(s) + I_2(aq) e. Pb(s) + e^- + 2I^-(aq) rightarrow PbI_2(aq) Write a balanced chemical equation for the overall reaction represented below. AI| AI^3+ || Zn | Zn^2+ a. 3Zn(s) + 2AI(s) rightarrow 2AI^+(aq) + 3Zn^2+(aq) b. Zn(s) + 2AI^+(aq) rightarrow 2AI(s) + Zn^2+(aq) c. 3Zn^2+(aq) + 2AI^3+(aq) rightarrow 2AI(s) + 3Zn(s) d. 3Zn^2+(aq) + 2AI(s) rightarrow 2AI^3+(aq) + 3Zn(s) e. Zn(s) + Al^3+(aq) rightarrow AI(s) + Zn^2+(aq) Write a balanced chemical equation for the overall reaction represented below. Pt | Fe^2+, Fe^3+ || CI^- | AgCI | Ag a. AgCI(s) + Fe^3+(aq) + 2e^- rightarrow Ag(s) + Fe^2+(aq) + Cl^-(aq) b. AgCI(s) + Fe^3+(aq) rightarrow Ag(s) + Fe^2+(aq) + CI^-(aq) c. AgCI(s) + Fe^2+(aq) rightarrow Ag(s) + Fe^3+(aq) + Cl^-(aq) d. Ag(s) + Fe^3+(aq) + CI^-(aq) rightarrow AgCI(s) + Fe^2t(aq) e. Ag(s) + Fe^2+(aq) + CI^-(aq) rightarrow AgCI(s) + Fe^3+(aq) Write a balanced chemical equation for the overall reaction represented below. Cu | Cu^2+ || CI^- | Hg_2CI_2 | Hg a. 2Hg(l) + Cu^2+(aq) + 2CI-(aq) rightarrow Hg_2Cl_2(s) + Cu(s) b. Hg_2CI_2(s) + Cu(s) rightarrow 2Hg(l) + Cu^2+(aq) + 2CI^-(aq) c. Hg_2CI_2(s) rightarrow 2Hg(l) + Cl_2(aq) d. Hg_2Cl_2(s) + Cu(s) rightarrow 2Hg(l) + Cu^2+(aq) + CI_2(aq) e. 2Hg(l) + Cl_2(aq) rightarrow Hg_2CI_2(s) What is the correct cell notation for the reaction below? Cd^2+(aq) + Ni(j) rightarrow Cd(s) + Ni^2+(aq) a. Cd^2+ | Cd || Ni | Ni^2+ b. Cd | Cd^2+ || Ni^2+ | Ni c. Ni | Ni^2+ || Cd^2+ | Cd d. Ni | Cd2^|| Ni^2+ | Cd e. Cd^2+ | Cd || Ni | Ni^2+Explanation / Answer
1. The cell diagram is a shorthand notation to represent the redox reactions of an electrical cell.
2.A double vertical line (||) is used to separate the anode half reaction from the cathode half reaction. This represents the salt bridge.
3.The anode (where oxidation occurs) is placed on the left side of the ||.
4.The cathode (where reduction occurs) is placed on the right side of the ||.
5.A single vertical line (|) is used to separate different states of matter on the same side, and a comma is used to separate like states of matter on the same side.
20. Pb(s) + I2(aq) ------> PbI2(s) is the net balanced equation.
option b.
Oxidation: Pb(s) ---------> Pb+2 (aq) +2e-
Reduction: I2(aq) + 2e- -------------> 2I- (aq)
Net: Pb(s) + I2(aq) ------> PbI2(s)
21. SOLUTION
Oxidation: {Al(s) Al3+(aq) +3e-} x 2
Reduction: {Zn2+(aq) +2e- Zn(s)} x 3
Net: 2Al(s) + 3Zn2+(aq) 2Al3+(aq) + 3Zn(s)
22. Oxidation: Fe+2 (aq) -----------> Fe+3 (aq) +e-
Reduction: AgCl(s) + e- ---------------------> Ag(s) + Cl- (aq)
Net: AgCl(s) + Fe+2 (aq) --------------> Ag(s) + Cl- (aq) + Fe+3 (aq)
23.Oxidation:Cu (s) --> Cu2+(aq) + 2 e-
Reduction: Hg2Cl2(s) + 2e- -----------> 2Hg (l) + 2Cl-(aq)
Net: Hg2Cl2(s) + Cu (s) ---------------> Cu2+(aq) + 2Hg (l) + 2Cl-(aq)
24. The anode (where oxidation occurs) is placed on the left side of the ||.
The cathode (where reduction occurs) is placed on the right side of the ||.
A single vertical line (|) is used to separate different states of matter on the same side, and a comma is used to separate like states of matter on the same side.
Ni/Ni+2 // Cd+2/Cd is the answer.
Because Ni is undergoing oxidation and Cadmium is undergoing reduction.
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