Homework Answers
Correct answer : a) 1.24 V
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Half-Reaction Fe2+(aq) + 2e Fe(s) Hg2+ (aq) + 2e Hg() Ag+ (aq) + e + Ag(s)...
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...
Please combine the Mn electrode half reaction with the Ag electrode half reaction and write the...
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...
Fill in the Blanks 0.34 Half Reaction E. (V) Half Reaction Erd® (V) F + 2e...
Fill in the Blanks 0.34 Half Reaction E. (V) Half Reaction Erd® (V) F + 2e →2F 0+ 2H + 2e →H,02 0.68 Ag* + e → Ag Cu* + e → Cu 0.52 Co? + e Co? O2 + 2H,0 + 4e → 40H 0.40 H,O, + 2H + 2e → 2H,0 Cu2+ + 2e → Cu Cet+e → Ces Cu2+ + e → Cut 0.16 PbO, + 4H+ +50,2 +2e → PbSO, + 2H,0 2H* + 2e →...
Given: Zn2+ (aq) + 2e Zn(s); E--0.76 V Cu²+ (aq) +20 Cu(s); E° -0.34 V What...
Given: Zn2+ (aq) + 2e Zn(s); E--0.76 V Cu²+ (aq) +20 Cu(s); E° -0.34 V What is the cell potential of the following electrochemical cell at 25°C? Zn(s) | Zn2+(1.0 M) || Cu2+(0.0010 M) Cu(s) a) greater than 1.10 v b) between 0.76 and 1.10 V c) less than 0.42 V d) between 0.00 and 0.76 V e) between 0.34 and 0.76 V
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 =...
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 +...
Find the best combination of half-cell pair from the following list, which will give the highest...
Find the best combination of half-cell pair from the following list, which will give the highest voltage. What is the voltage for that Galvanic cell? Given that Reduction Half-reaction Standard Potential (Eredo) Zn2+(aq) + 2e– → Zn(s) -0.763 (V) Fe2+(aq) + 2e– → Fe(s) -0.44 (V) Cu2+(aq) + 2e– → Cu(s) +0.34 (V) Sn2+(aq) + 2e– → Sn(s) -0.14 (V) Cu2+(aq) + e– → Cu+(aq) + 0.153 (V) Ag+(aq) + e– → Ag(s) + 0.80 (V) Cu+(aq) + e– →...
2.87 Ered® (V) 0.68 0.52 0.40 0.34 0.16 Half Reaction F,+ 2e →2F Ag* + e...
2.87 Ered® (V) 0.68 0.52 0.40 0.34 0.16 Half Reaction F,+ 2e →2F Ag* + e → Ag Co3 + e + CO2- H2O2 + 2H+ + 2e → 2H,0 Ce4+ + e → Ce+ PbO, + 4H+ + SO42- + 2e → PbSO, + 2H,0 Mno, + 4H+ + 3e → MnO2 + 2H,0 2e + 2H+ + 10, → 103 + H2O Mn0, +8H+ + 5e → Mn2+ + 4H,0 Aul+ + 3e → Au Cl2 + 2e...
Using the standard reduction potentials given below, choose the reaction than can only be achieved through...
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) +...
What is the value of Δ G° (in kJ ) for the following balanced redox reaction at 25.0℃ ?
Half-ReactionE°(V)Fe^(2+)(aq)+2e^(-)rarrFe(s)-0.44Hg^(2+)(aq)+2e^(-)rarrHg(l)0.86Ag^(+)(aq)+e^(-)rarrAg(s)0.80Cu^(2+)(aq)+2e^(-)rarrCu(s)0.34Zn^(2+)(aq)+2e^(-)rarrZn(s)-0.76What is the value of Δ G° (in kJ ) for the following balanced redox reaction at 25.0℃ ?Hg²⁺(a q)+Fe(s) ⟶ Hg(l)+Fe²⁺(a q)a) -40.5 kJb) -81.1 kJc) -125 kJd) -251 kJ
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...
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...
Fill in the Blanks 0.34 Half Reaction E. (V) Half Reaction Erd® (V) F + 2e →2F 0+ 2H + 2e →H,02 0.68 Ag* + e → Ag Cu* + e → Cu 0.52 Co? + e Co? O2 + 2H,0 + 4e → 40H 0.40 H,O, + 2H + 2e → 2H,0 Cu2+ + 2e → Cu Cet+e → Ces Cu2+ + e → Cut 0.16 PbO, + 4H+ +50,2 +2e → PbSO, + 2H,0 2H* + 2e →...
Given: Zn2+ (aq) + 2e Zn(s); E--0.76 V Cu²+ (aq) +20 Cu(s); E° -0.34 V What is the cell potential of the following electrochemical cell at 25°C? Zn(s) | Zn2+(1.0 M) || Cu2+(0.0010 M) Cu(s) a) greater than 1.10 v b) between 0.76 and 1.10 V c) less than 0.42 V d) between 0.00 and 0.76 V e) between 0.34 and 0.76 V
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 =...
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 +...
2.87 Ered® (V) 0.68 0.52 0.40 0.34 0.16 Half Reaction F,+ 2e →2F Ag* + e → Ag Co3 + e + CO2- H2O2 + 2H+ + 2e → 2H,0 Ce4+ + e → Ce+ PbO, + 4H+ + SO42- + 2e → PbSO, + 2H,0 Mno, + 4H+ + 3e → MnO2 + 2H,0 2e + 2H+ + 10, → 103 + H2O Mn0, +8H+ + 5e → Mn2+ + 4H,0 Aul+ + 3e → Au Cl2 + 2e...
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) +...
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