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Design a voltaic cell with the following two reduction half -reactions : Design a voltaic cell...
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.
What is the anode of the voltaic cell made by the combination of following half reactions? Reaction Standard Reduction Potential Zn2+ + 2e– → Zn E0 = –0.76 V Ag+ + e– → Ag E0 = +0.80 V Question options: a. Zn2+(aq) b. Ag+(aq) c. Ag(s) d. Zn(s)
Consider the two reduction half-reactions: Br2(l) + 2 e− ⟶ 2 Br−(aq) Eo = 1.09 V Ag+(aq) + e− ⟶ Ag(s) Eo = 0.337 V Use the electrode potentials above to calculate Eocell and ∆Gorxn for the reaction below, and determine if it is the reaction for a voltaic cell or an electrolytic cell.
In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2 half- cell and an H2/H half-cell under the following conditions: [Zn2] = 0.042 M [H]- 19 M partial pressure of H2 =0.37 atm. Calculate Ecell at 298 K (enter to 3 decimal places). Zn2 (aq) + 2e +2H (aq) + 2e1 Eo-0.76 V E 0.00 V Zn(s) H2(g)
Use the half-reactions below to produce a voltaic cell with the given standard cell potential. Standard Cell Potential Co- (aq) + e-Cot (aq) E = +1.82 V 1.53 V 2H(aq) + 2e-H2(g) E = +0.00 V Pb2+ (aq) + 2e-Pb(s) E = -0.13 V Fe (aq) + e-Fel+ (aq) E = +0.77 V Ag (aq) + e-Ag(s) E = +0.80 V Sn* (aq) + 2e Sne (aq) 20.13 V Cu- (aq) + e- Cu(aq) E = +0.15 V Zn²+ (aq)...
A battery is constructed at 25ºC using the voltaic cell with initial concentrations Zn│Zn2+ (0.100 M) ││Ag+ (1.500 M) │Ag How much does the cell voltage drop when 95% of the capacity (i.e., the concentration of Ag+ drops to 5% of its starting value) is consumed? The standard reduction potentials for the two half-cells are + Zn2+ + 2e- → Zn Eº = − 0.76 V Ag+ + e- → Ag Eº = 0.80 V
A battery is constructed at 25ºC using the voltaic cell with initial concentrations Zn│Zn2+ (0.100 M) ││Ag+ (1.500 M) │Ag How much does the cell voltage drop when 95% of the capacity (i.e., the concentration of Ag+ drops to 5% of its starting value) is consumed? The standard reduction potentials for the two half-cells are Zn2+ + 2e- → Zn Eº = − 0.76 V Ag+ + e- → Ag Eº = 0.80 V
Calculate the theoretical cell potential (E°) of a galvanic cell under standard conditions made up of copper and magnesium (see Part II and Table 1 for more information). PARTIL Creating and Testing Voltaic Cells Introduction and Background for the Voltaic Cells A galvanic cell (sometimes more appropriately called a voltaic cell) consists of two half-cells joined by a salt bridge that allow ions to pass between the two sides in order to maintain electroneutrality. Each half-cell contains the Components of...
In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions: [Zn2+ ] = 0.021 M [H+ ]= 1.3 M partial pressure of H2 = 0.32 atm. Calculate Ecell at 298 K (enter to 3 decimal places). Zn2+ (aq) + 2e − ⟶ Zn(s) E° = − 0.76 V 2H+ (aq) + 2e − ⟶ H2(g) E° = 0.00 V We were unable...
Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e– → Sn(s). What is the concentration of Sn2+ if Zn2+ is 2.5 × 10−3 M and the cell emf is 0.660 V? The standard reduction potentials are given below Zn+2(aq) + 2 e− → Zn(s) E∘red == −0.76 V Sn2+(aq) + 2 e– → Sn(s) E∘red −0.136 V