A voltaic cell is based on the reduction of Ag+(aq) to Ag(s) and the oxidation of Sn(s) to Sn2+(aq).
a) Draw the cell diagram
A voltaic cell is based on the reduction of Ag+(aq) to Ag(s) and the oxidation of...
A voltaic cell is based on the reduction of Ag^+(aq) to Ag(s) and the oxidation of Sn(s) to Sn^2+(aq). (a) Write half-reactions for the cell's anode and cathode. Include the phases of all species in the chemical equation. Anode Cathode (b) Write a balanced cell reaction. Include the phases of all species in the chemical equation.
QUESTION 13 Consider a voltaic cell based on the half-cells: Ag (aq) + e - Ag(s) E = +0.80 V Sn2+ (aq) + 2 e Sn(s) E = -0.14 V Identify the anode and give the cell voltage under standard conditions: O A Sn: Eºcell = -0.66 V B. Sn; Eºcell = 0.66 V O C. Ag: Eºcell = 0.67 V D. Ag: Eºcell = 0.94 V E. Sn; Eºcell = 0.94 V
A voltaic cell is based on the following two half-reactions: Cd2 (ag) +2e-> Cd (s) Sn2(aq)+ 2e Sn (s) Calculate the standard cell potential. Use the date from the attached table.SRP2.docx Oa 0.13 Ob 042 Oc.027 Od-0.27
Design a voltaic cell with the following two reduction half-reactions: Ag+(aq) + e− ⟶ Ag(s) Eo = 0.80 V Pb2+(aq) + 2 e− ⟶ Pb(s) Eo = −0.13 V Calculate Eocell and the equilibrium constant K for the voltaic cell at 298 K. Click here for a copy of Final Exam cover sheet.
A voltaic cell is made from Ni(s), Ni2+(aq), Ag(s) and Ag+ a. Fill in the diagram with the chemicals associated with each part of the cell. b. Write a balanced cell reaction A voltaic cell is made from Ni(s), Niz*(aq), Ag(s) and Ag 7. a. Fill in the diagram with the chemicals associated with each part of the cell. Ni2+ + 2 e-→ Ni(s) Ag+ + e-→ Ag(s) -0.25 0.80
5. A voltaic cell is constructed based on the oxidation of zinc metal and the reduction of silver (I) ions. Solutions of 1.00 M silver nitrate and zinc nitrate also were used. The anode is on the left, the cathode is on the right. Where does oxidation occur? 6. For each of the numbered bonds in the figure below, identify whether the bond is cis, trans, or neither cis trans. A) I-cis; II-trans, III-cis C) I-neither, II-trans; IIl-cis D) I-neither,...
A voltaic cell is constructed that is based on the following reaction: Sn2+(aq)+Pb(s)→Sn(s)+Pb2+(aq). a. If the concentration of Sn2+ in the cathode compartment is 1.50 M and the cell generates an emf of 0.22 V , what is the concentration of Pb2+ in the anode compartment? b. If the anode compartment contains [SO2−4]= 1.50 M in equilibrium with PbSO4(s), what is the Kspof PbSO4?
Given the following standard reduction potentials choose the cell which will work as a voltaic cell. All cells below are written according to the usual cell diagram convention. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V 2H+(aq) + 2e → H2(g) E° = 0.00 V Sn2+ (aq) + 2e → Sn(s) E° = -0.14 V Ni2+(aq) + 2e → Ni(s) E° = -0.26 V Cd2+(aq) + 2e → → Cd(s) E° = -0.40 V Sn(s) | Sn2+(aq) || Ni2+(aq)...
An electrochemical cell is based on the following two half-reactions:oxidation: Sn(s)→Sn2+(aq,1.60 M) +2e-reduction: ClO2(g, 0.130 atm )+e−→ClO2-(aq, 1.55 MM )Compute the cell potential at 25°C
Consider a voltaic cell based on the half-cells: Ag+ (aq) + e - Ag(s) E* = +0.80 V Pb2+(aq) + 2 e-Pb(s) E* = -0.13 V Identify the anode and give the cell voltage under standard conditions: = 0.67 V Ag: E cell B. Pb: E cell = 0.67 V OC. Pb: E cell = 0.93 V D. Pb: E cell = -0.67 V E. Ag: E cell = 0.93 V