Instead of using zinc, a student accidentally used a 1 M solution of silver (Ag). Based...
Instead of using zinc, a student accidentally used a 1 M solution of silver (Ag). Based on the information in the table below, what would be the theoretical cell voltage, Eocell, when it is used in a galvanic cell with copper?
| Electrode | Eo |
| Ag+ + e- à Ag | +0.80 V |
| Cu2+ + 2e- à Cu | +0.34 V |
| Pb2+ + 2e- à Pb | -0.13 V |
| Zn2+ + 2e- à Zn | -0.76 V |
| Al3+ + 3e- à Al | -1.66 V |
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Instead of using zinc, a student accidentally used a 1 M
solution of silver (Ag). Based...
1) Instead of using zinc, a student accidentally used a 1 M solution of silver (Ag)....
1) Instead of using zinc, a student accidentally used a 1 M solution of silver (Ag). Based on the information in the table below, what would be the theoretical cell voltage, Eocell, when it is used in a galvanic cell with copper?
Q. Consider a galvanic cell with a zinc electrode immersed in 1.0M Zn2+ and a silver...
Q. Consider a galvanic cell with a zinc electrode immersed in 1.0M Zn2+ and a silver electrode immersed in 1.0M Ag+. Which of the electrodes is the anode? Zn2 + 2e- --> Zn E° = -0.76 V Ag+ + e- --> Ag E° = 0.80 V
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...
What is the standard emf of a galvanic cell made of a Co electrode in a 1.0 M Co(NO32 solution and a Al electrode in a 1.0 M AI(NO3)3 solution at 25°C?
What is the standard emf of a galvanic cell made of a Co electrode in a 1.0 M Co(NO32 solution and a Al electrode in a 1.0 M AI(NO3)3 solution at 25°C? 0 cell Standard Reduction Potentials at 25°C Half-Reaction E(V +2.87 +2.07 +1.82 O,(g) 2H (aq)2e0(g)+HO Co3+(aq) + e-_? Co2+(aq) H,02(aq) + 2H"(aq) + 2e-_ 2H20 Cu2+(aq) + 2e-? Cu(s) AgCIs) + Ag(s) + CI(a) S02-(aq) + 4H'(aq) + 2e S02(g) + 2H20 Cu2+(aq) + e-_ Cu+(aq) Sn (aq)...
PART A: REDOX REACTIONS 1. For each of the metals, write the redox equations for reactions...
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...
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 +...
Using the information in the table: Which combination of metals, if used to create an electrochemical...
Using the information in the table:
Which combination of metals, if used to create an
electrochemical cell, would produce the largest voltage?
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 + Cu2+(aq) + 2e → Cu(s) Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) E half-cell = 0.7628 V Eºhalf-cell =...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...
19 20 Question 16 Half-cell Potentials: Half Reaction: E' value + 0.80 V +0.77 V Agte...
19 20 Question 16 Half-cell Potentials: Half Reaction: E' value + 0.80 V +0.77 V Agte → AS Fe3+ + + Fe2+ Cu2+ 2e → Cu Pb2+ + 2e → Pb +0.34 V -0.13 V Ni2+ + 2e → NI -0.25 V - 0.40 V Cd2+ +2e → ca Fe2+ + 2e → Fe Zn2+ + 2e → Zn -0.44 V - 0.76 V A13+ +3 → AI - 1.66 V Consider an electrochemical cell constructed from the following half...
Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g)...
Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g) Express your answer using two significant figures. Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V) Half-reaction E∘ (V) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) −0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) −0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) −0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) −0.76 Cu2+(aq)+2e−→Cu(s) 0.34 Mn2+(aq)+2e−→Mn(s) −1.18 2H+(aq)+2e−→H2(g) 0.00 Al3+(aq)+3e−→Al(s) −1.66 Fe3+(aq)+3e−→Fe(s) −0.036 Mg2+(aq)+2e−→Mg(s) −2.37 Pb2+(aq)+2e−→Pb(s) −0.13 Na+(aq)+e−→Na(s) −2.71 Sn2+(aq)+2e−→Sn(s) −0.14 Ca2+(aq)+2e−→Ca(s) −2.76 Ni2+(aq)+2e−→Ni(s) −0.23 Ba2+(aq)+2e−→Ba(s) −2.90 Co2+(aq)+2e−→Co(s) −0.28 K+(aq)+e−→K(s) −2.92 Cd2+(aq)+2e−→Cd(s)...
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...
What is the standard emf of a galvanic cell made of a Co electrode in a 1.0 M Co(NO32 solution and a Al electrode in a 1.0 M AI(NO3)3 solution at 25°C? 0 cell Standard Reduction Potentials at 25°C Half-Reaction E(V +2.87 +2.07 +1.82 O,(g) 2H (aq)2e0(g)+HO Co3+(aq) + e-_? Co2+(aq) H,02(aq) + 2H"(aq) + 2e-_ 2H20 Cu2+(aq) + 2e-? Cu(s) AgCIs) + Ag(s) + CI(a) S02-(aq) + 4H'(aq) + 2e S02(g) + 2H20 Cu2+(aq) + e-_ Cu+(aq) Sn (aq)...
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 +...
Using the information in the table:
Which combination of metals, if used to create an
electrochemical cell, would produce the largest voltage?
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 + Cu2+(aq) + 2e → Cu(s) Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) E half-cell = 0.7628 V Eºhalf-cell =...
19 20 Question 16 Half-cell Potentials: Half Reaction: E' value + 0.80 V +0.77 V Agte → AS Fe3+ + + Fe2+ Cu2+ 2e → Cu Pb2+ + 2e → Pb +0.34 V -0.13 V Ni2+ + 2e → NI -0.25 V - 0.40 V Cd2+ +2e → ca Fe2+ + 2e → Fe Zn2+ + 2e → Zn -0.44 V - 0.76 V A13+ +3 → AI - 1.66 V Consider an electrochemical cell constructed from the following half...
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