Part I: Voltaic Cells Use your textbook or other resource to complete the following table: Half-Reaction...
predicted voltage 1,2,3? ulations Part I: Voltaic Cells A. Cell Potentials Using a Zn(s)/Zn (aq) Reference Electrode Complete the following table using your mean values for PartI A, and assigning a reference voltage of 0.00 volts for the reduction of Zn2 to Zn. Put the voltages in increasing order Half-Reaction Standard Reduction Potential at 25 °C, Volts Zn (aq) + 2e Zn (s) 0.00 Fers +7 0.30 0.5 Phess B.Cell Potentials Using the Table Above Redox Couple predicted voltage observed...
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...
Table view List view Ecell (calculated) Ecell (calculated) 1.115 1.124 Table 1. Voltaic cells data table Ecell (measured) Reaction Quotient Q 1. Zn | Zn2(1.0M) || Cu2+(1.0 M) | Cu 2. Zn | Zn2+(1.0 M) || Cu2+(0.1 M) Cu 1.067 3. ZnZn2+(0.1 M) || Cu2+(1.0 M) | Cu 4. Zn | Zn2+(1.0 M) || Pb2+(1.0 M) | Pb 5. ZnZn2(1.0M) || Pb2+(0.1 M) | Pb 0.600 6. Zn | Zn2+(0.1 M) || Pb2+(1.0 M) I Pb 0.644 7. ZnZn2+(1.0 M)...
Please combine the Mn electrode half reaction with the Ag electrode half reaction and write the complete redox reaction using date from table 8.1. Indicate which is the oxidizing and which is the reducing agent. b. Calculate the EMF when the reaction quotient (Q) = ]Mn2+]/[Ag+]^2 = 10^-5 Reducing agent Half-reaction E0 (volts) -2.93 2.87 Ca Na Mg Mn Zn Fe Ni Pb Ca Ca2++2e Na → Na++e- -2.36 Mn Mn++2e Zn Zn2++2e Fe → Fe2++ 2e- NiNi2++2e- -0.76 -0.47...
PART A: REDOX REACTIONS 1. For each of the metals, write the redox equations for reactions you observed in a table as shown below. Write NR for “no reaction” where none was observed. 2+ (**e.g., Cu + Zn → Cu + Zn , Ecell = 1.10 V) Cu(NO3)2 Pb(NO3)2 Zn(NO3)2 16 Cu(s) NR NR Pb(s) NR Zn(s) NR 2. Calculate the Eº for every cell, whether or not a reaction was observed, using equation (5) and values for the standard...
Using the table below: 19. Three combinations of metals are listed below, which combination would produce the largest voltage if they were used to construct an electrochemical cell? Copper (Cu) with zinc (Zn) Lead (Pb) with zinc (Zn) Lead (Pb) with cadmium (Cd) Liu lur the reaction between Zn and Cu2+ ions is 1.1030 V, we can use the known value for the half-cell potential for zinc to determine the half-cell potential for copper: Zn(s) → Zn2+(aq) + 2e +...
The following half-cells are availble: (i) Ag+ (1.0 M) ∣ Ag (s) 0.7994 v (ii) Zn2+ (1.0 M) ∣ Zn (s) -0.76 v (iii) Cu2+ (1.0 M) ∣ Cu (s) 0.337 v (iv) Co2+ (1.0 M) ∣ Co (s) -0.28 v Linking any two half-cells makes a voltaic cell. Given four different half-cells, six voltaic cells are possible. These are labeled, for simplicity, Ag - Zn, Ag-Cu, Ag-Co, Zn - Cu, Zn-Co, and Cu-Co. (the first metal is the anode...
Please help, thanks. Use the table below to answer the questions that follow. Half Reaction E°(V) Ag+ + e- → Ag (s) +0.80 Cu2+ + 2e- → Cu (s) +0.34 Pb2+ + 2e- → Pb (s) -0.13 Ni2+ + 2e- → Ni (s) -0.28 (A) Calculate E°cell for a voltaic cell made with Ni, Ni2+and Ag, and Ag+. (B) Which substance is Oxidized? (be sure to include the state of the oxidized substance in your answer) Zn(s) + CuSO4(aq) -->...
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...
Question 3 6.66 pts The following half-cells are availble: (i) Ag (1.0M) 1 Ag (s) (ii) Zn2+ (1.0 M) | Zn (s) (iii) Cu2+ (1.0 M)Cu (s) (iv) Co2+ (1.0 M)Co (s) 0.7994 v -0.76 V 0.337 v -0.28 v Linking any two half-cells makes a voltaic cell. Given four different half-cells, six voltaic cells are possible. These are labeled, for simplicity, Ag - Zn, Ag-Cu, Ag-Co, Zn - Cu, Zn-Co, and Cu-Co. (the first metal is the anode and...