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V-25 OBSERVATIONS SHEET - EXPERIMENT V Voltaic Cell and the Nernst Equation Part I: Electrode Potentials...
In a copper-zinc voltaic cell, one half-cell consists of a ZnZn electrode inserted in a solution...
In a copper-zinc voltaic cell, one half-cell consists of a ZnZn
electrode inserted in a solution of zinc sulfate and the other
half-cell consists of a CuCu electrode inserted in a copper sulfate
solution. These two half-cells are separated by a salt bridge.
At the zinc electrode (anode), ZnZn metal undergoes oxidation by
losing two electrons and enters the solution as Zn2+Zn2+ ions. The
oxidation half-cell reaction that takes place at the anode is
Zn(s)→Zn2+(aq)+2e−Zn(s)→Zn2+(aq)+2e−
The CuCu ions undergo reduction...
Use the table of Standdard Reduction Potentials and the Nernst Equation to calculate the concentration of...
Use the table of Standdard Reduction Potentials and the Nernst
Equation to calculate the concentration of Cu2+ present in the
Cu/Cu2+ half-cell after the addition NH3 (aq) while it was coupled
with
a) Zn/Zn2+
b) Ag/Ag+
Table 2 has your measured voltages for these cells. SHOW YOUR
SETUP OF THE NERST EQUATION WITH ALL VARIABLES FILLED IN. SHOW THE
CALCULATED [Cu2+] CONCENTRATION.
Table 2:
METAL
Mg
Ag
Ni
Zn
Pb
Cu
1.230 V
0.642 V
0.030 V
0.648 V
0.256...
20. A galvanic (voltaic) cell is constructed froma half-cell containing a solid aluminum electrode dipped in...
20. A galvanic (voltaic) cell is constructed froma half-cell containing a solid aluminum electrode dipped in a 1.0 M Al(NO3)3 solution and a half-cell containing a solid indium electrode in contact with a 1.0 M In(NO3)3 solution. The half-cells are linked by an external circuit (a wire) and by a KCl salt bridge. The cell generates a standard voltage of 1.32 V and the indium electrode is positive. The standard half-cell reduction potential for aluminum, E°(Al3*/Al), is listed in the...
predicted voltage 1,2,3? ulations Part I: Voltaic Cells A. Cell Potentials Using a Zn(s)/Zn (aq) Reference...
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...
2 A voltaic cell is set up with one beaker containing 1.0 M Cu(NO 3) 2...
2 A voltaic cell is set up with one beaker containing 1.0 M Cu(NO 3) 2 and a copper electrode, and another beaker containing 1.0 M Mn(NO 3) and a manganese electrode. Given the following standard reduction potentials, answer the 3 questions below: E Cu2+(aq) + 2e Cu(s) 0.34 V Mn2+(aq) + 2e + Mn(s) 1.18V Part a. Write out the half-cell reaction that occurs at the anode of the voltaic cell. Part b. In which direction do electrons flow?...
Half-cell Potentials: Half Reaction: E value +0.80 V Agt + e → Ag Fe3+ + €...
Half-cell Potentials: Half Reaction: E value +0.80 V Agt + e → Ag Fe3+ + € → Fe2+ +0.77 v +0.34 V -0.13 V Cu2+ +2e → Cu Pb2+ + 2e - → Ib Ni2+ + 2e → Ni Cd2+ +2e → Cd -0.25 V -0.40 V Fe2+ + 2e → Fe -0.44 V Zn2+ + 2e → Zn -0.76 V Al3+ +3e → AI - 1.66 V Consider an electrochemical cell constructed from the following half cells, linked by...
Lab11_PP_147-156... * 5 79 Calculations Part I:Voltaic Cells A. Cell Potentials Using a Zn(s)[Zn2+ (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...
salt bridge Zn(s) electrode Culs) electrode 1.0M Zn (a 1.0 M Cu (aq A voltaic cell...
salt bridge Zn(s) electrode Culs) electrode 1.0M Zn (a 1.0 M Cu (aq A voltaic cell similar to that shown in the figure above is constructed. The electronic device shown at the top of the figure is a volt meter. One electrode compartment consists of a zinc strip placed in a 1.0 M ZnCl2 solution, and the other has a copper strip placed in a 1.0 M CuSOA solution. The overall cell reaction is: Zn(s)Cu2+(aq)= zn2+ (aq ) Cu (s)...
how do I solve this? 1. Use the Nernst equation to calculate the standard cell potential...
how
do I solve this?
1. Use the Nernst equation to calculate the standard cell potential for a voltaic cell of a Sn electrode in 0.10M Sn2(a) in one half-cell and Al in 0.10M AP) in the other. (EPe Eaode- Emode 1.52 V, Show your work. Hint: Check E values of the reduction haif reactions. Write the one with more negative E as oxidation half reaction (change the sign); ensure that equal number of electrons are exchanged ie. electrons cancel,...
38. The following redox half reactions are combined in a voltaic cell. Which reaction occurs at...
38. The following redox half reactions are combined in a voltaic cell. Which reaction occurs at the cathode and what is the Eceu? Fe2+(aq) + 2e → Fe(s) E°=-0.44 V Cu²+(aq) + 2e → Cu(s) E°= 0.34 V a) b) c) d) Cu2+(aq) + 2e → Cu(s), Ecell = 0.78 V Fe2+(aq) + 2e → Fe(s), Ecel = 0.78 V Fe2+(aq) + 2e → Fe(s), Ecell =-0.10 V Cu²+(aq) + 2e → Cu(s), Ecel = 0.10 V Cu²+ (aq) +...
In a copper-zinc voltaic cell, one half-cell consists of a ZnZn
electrode inserted in a solution of zinc sulfate and the other
half-cell consists of a CuCu electrode inserted in a copper sulfate
solution. These two half-cells are separated by a salt bridge.
At the zinc electrode (anode), ZnZn metal undergoes oxidation by
losing two electrons and enters the solution as Zn2+Zn2+ ions. The
oxidation half-cell reaction that takes place at the anode is
Zn(s)→Zn2+(aq)+2e−Zn(s)→Zn2+(aq)+2e−
The CuCu ions undergo reduction...
Use the table of Standdard Reduction Potentials and the Nernst
Equation to calculate the concentration of Cu2+ present in the
Cu/Cu2+ half-cell after the addition NH3 (aq) while it was coupled
with
a) Zn/Zn2+
b) Ag/Ag+
Table 2 has your measured voltages for these cells. SHOW YOUR
SETUP OF THE NERST EQUATION WITH ALL VARIABLES FILLED IN. SHOW THE
CALCULATED [Cu2+] CONCENTRATION.
Table 2:
METAL
Mg
Ag
Ni
Zn
Pb
Cu
1.230 V
0.642 V
0.030 V
0.648 V
0.256...
20. A galvanic (voltaic) cell is constructed froma half-cell containing a solid aluminum electrode dipped in a 1.0 M Al(NO3)3 solution and a half-cell containing a solid indium electrode in contact with a 1.0 M In(NO3)3 solution. The half-cells are linked by an external circuit (a wire) and by a KCl salt bridge. The cell generates a standard voltage of 1.32 V and the indium electrode is positive. The standard half-cell reduction potential for aluminum, E°(Al3*/Al), is listed in the...
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...
2 A voltaic cell is set up with one beaker containing 1.0 M Cu(NO 3) 2 and a copper electrode, and another beaker containing 1.0 M Mn(NO 3) and a manganese electrode. Given the following standard reduction potentials, answer the 3 questions below: E Cu2+(aq) + 2e Cu(s) 0.34 V Mn2+(aq) + 2e + Mn(s) 1.18V Part a. Write out the half-cell reaction that occurs at the anode of the voltaic cell. Part b. In which direction do electrons flow?...
Half-cell Potentials: Half Reaction: E value +0.80 V Agt + e → Ag Fe3+ + € → Fe2+ +0.77 v +0.34 V -0.13 V Cu2+ +2e → Cu Pb2+ + 2e - → Ib Ni2+ + 2e → Ni Cd2+ +2e → Cd -0.25 V -0.40 V Fe2+ + 2e → Fe -0.44 V Zn2+ + 2e → Zn -0.76 V Al3+ +3e → AI - 1.66 V Consider an electrochemical cell constructed from the following half cells, linked by...
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...
salt bridge Zn(s) electrode Culs) electrode 1.0M Zn (a 1.0 M Cu (aq A voltaic cell similar to that shown in the figure above is constructed. The electronic device shown at the top of the figure is a volt meter. One electrode compartment consists of a zinc strip placed in a 1.0 M ZnCl2 solution, and the other has a copper strip placed in a 1.0 M CuSOA solution. The overall cell reaction is: Zn(s)Cu2+(aq)= zn2+ (aq ) Cu (s)...
how
do I solve this?
1. Use the Nernst equation to calculate the standard cell potential for a voltaic cell of a Sn electrode in 0.10M Sn2(a) in one half-cell and Al in 0.10M AP) in the other. (EPe Eaode- Emode 1.52 V, Show your work. Hint: Check E values of the reduction haif reactions. Write the one with more negative E as oxidation half reaction (change the sign); ensure that equal number of electrons are exchanged ie. electrons cancel,...
38. The following redox half reactions are combined in a voltaic cell. Which reaction occurs at the cathode and what is the Eceu? Fe2+(aq) + 2e → Fe(s) E°=-0.44 V Cu²+(aq) + 2e → Cu(s) E°= 0.34 V a) b) c) d) Cu2+(aq) + 2e → Cu(s), Ecell = 0.78 V Fe2+(aq) + 2e → Fe(s), Ecel = 0.78 V Fe2+(aq) + 2e → Fe(s), Ecell =-0.10 V Cu²+(aq) + 2e → Cu(s), Ecel = 0.10 V Cu²+ (aq) +...
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