predicted voltage 1,2,3? ulations Part I: Voltaic Cells A. Cell Potentials Using a Zn(s)/Zn (aq) Reference...
Lab11_PP_147-156... * 5 79 Calculations Part I:Voltaic Cells A. Cell Potentials Using a Zn(s)[Zn2+ (aq) Reference Electrode Complete the following table using your mean values for Part 1 A, and assigning a reference voltage of 0.00 volts for the reduction of Zn?t to Zn. Put the voltages in increasing order. Half-Reaction Standard Reduction Potential at 25 °C, Volts Zn2+ (aq) + 2e → Zn (s) 0.00 0.328 0.443 0.907 observed voltage 0.469 percent error B. Cell Potentials Using the Table...
Part I: Voltaic Cells Use your textbook or other resource to complete the following table: Half-Reaction Zn2 ag)2eZn Fe (a)2e Fe (s) Standard Reduction Potential at 25 °C, Volts 0.76 13 34 Cu2+ (aq) +2e- Cu (s) → Use the table above to predict voltlages for the redox couples studied in Part L, sections A and B. Redox Couple predicted voltage observed voltage 1. Zn/Cu 2. Zn/Pb 3. Pb/Cu 4. Fe/Cu 87 405
Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2: Fe2+(aq)/Fe(s) and Zn2+(aq)/Zn(s) Which of the following is correct for the working cells? Standard reduction potentials, 298 K, Aqueous Solution (pH = 0): Cl2(g) + 2e --> 2C1-(aq); E° = +1.36 V Fe3+(aq) + e --> Fe2+(aq); E° = +0.77 V Cu2+(aq) + 2e --> Cu(s); E° = +0.34 V 2H+(aq) + 2e --> H2(g); E° = 0.00 V Pb2+(aq) + 2e --> Pb(s);...
Consider the voltaic cell and reduction half potentials: Zn(s) | Zn 2+(aq) (0.100 M) || Ni2+(aq) (1.50 M) | Ni(s) Zn 2+ (aq)/Zn (s) E o = - 0.760 V Ni2+ (aq)/Ni (s) E o = - 0.230 V a) Sketch the voltaic cell represented with the above line notation. Label the anode and cathode and indicate the half-reactions occurring at each electrode and the species present in each solution. Also indicate the direction of electron flow (3 marks). b)...
Using the standard reduction potentials given below, choose the reaction than can only be achieved through electrolysis. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V Pb2+(aq) + 2e + Pb(s) E° = -0.13 V Fe2+(aq) + 2e Fe(s) E° = -0.44 V Zn2+(aq) + 2e + Zn(s) E° = -0.77 V Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq) o Fe2+(aq) + Zn(s) → Fe(s) + Zn2+(aq) Pb2+(aq) + Fe(s) → Pb(s) + Fe2+(aq) Cu2+(aq) + Fe(s) → Cu(s) +...
Write the half reactions and overall reaction for each cell with calculated overall potentials as shown in Table 5-1. (Note: for the iron solutions the Nernst equation must be used) Pb(s) | Pb(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) Cu(s) Zn(s) | Zn(NO3)2(0.1M) || Cu(NO3)2 (0.1M) Cu(s) Cds) | Ca(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) | Cu(s) Cu() Cu(NO3)2(0.1M) Il Fe (0.1M/Fe? (0.1M graphite Pb(s) Pb(NO3)2(0.1M) Il Fe3(aq) (0.1M)/ Fe2(aq) (0.1MI graphite(s) Zns | Zn(NO3)2 (0.1M) || Pb(NO3)2 (0.1M) | Pb(s) Cdis Ca(NO3)2...
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
1. Some standard reduction potentials at 25 C are: E 0.337 V for Cu* (a) 2e- Cu(s) E" =-0.1 26 V for Phẩ (aq) + 2e → Pb(s) Eo-0.763 V for Zn2 (a)2e-> Zn(s) Predict the voltage of each of the following voltaic cells: a. Cu | Cu0.010 M)I Cu (0.010M) | Cu c. Zn | Zn2 (0010 M) Cu(0.010 M) | Cu b. Pb | Pb (0.010 M)1 Cu (0.010 MCud. Zn | Zn3 (0.010 M) 11 Pb (0.010...
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