Given the following standard half-cell potentials: MnO2(s) + 4H+ (aq) + 2e– → Mn2+(aq) + 2H2O(l)...
Given the following standard half-cell potentials:
MnO2(s) + 4H+ (aq) + 2e– → Mn2+(aq) + 2H2O(l) E° = 1.23 V
NO3 – (aq) + 4H+ (aq) + 3e– → NO(g) + 2H2O(l) E° = 0.96 V
N2(g) + 5H+ (aq) + 4e– → N2H5 + (aq) E° = –0.23 V
Which of the following reactions is nonspontaneous under standard state conditions?
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Given the following standard half-cell potentials:
MnO2(s) + 4H+ (aq) + 2e– → Mn2+(aq) + 2H2O(l)...
Standard Electrode Potentials at 25?C Reduction Half-Reaction E?(V) F2(g)+2e? ?2F?(aq) 2.87 Au3+(aq)+3e? ?Au(s) 1.50 Cl2(g)+2e? ?2Cl?(aq)...
Standard Electrode Potentials at 25?C Reduction Half-Reaction E?(V) F2(g)+2e? ?2F?(aq) 2.87 Au3+(aq)+3e? ?Au(s) 1.50 Cl2(g)+2e? ?2Cl?(aq) 1.36 O2(g)+4H+(aq)+4e? ?2H2O(l) 1.23 Br2(l)+2e? ?2Br?(aq) 1.09 NO3?(aq)+4H+(aq)+3e? ?NO(g)+2H2O(l) 0.96 Ag+(aq)+e? ?Ag(s) 0.80 I2(s)+2e? ?2I?(aq) 0.54 Cu2+(aq)+2e? ?Cu(s) 0.16 2H+(aq)+2e? ?H2(g) 0 Cr3+(aq)+3e? ?Cr(s) -0.73 2H2O(l)+2e? ?H2(g)+2OH?(aq) -0.83 Mn2+(aq)+2e? ?Mn(s) -1.18 How can the table be used to predict whether or not a metal will dissolve in HCl? In HNO3? Drag the terms on the left to the appropriate blanks on the right to...
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...
Half-potentials. Homework – Unanswered Due in 8 hours For the reaction: Au(s) + 4H" (aq) +...
Half-potentials. Homework – Unanswered Due in 8 hours For the reaction: Au(s) + 4H" (aq) + NO3(aq) + 4C"(aq) -- Aucla (aq) + NO(g) + 2H20 the standard potential E=-0.04 V. If the standard potential for the reduction of NO3 aq) to NO(g) is +0.96 V, determine the value of Efor the following half-reaction. AuCl(aq) + 3e - Au(s) + 4C1 (aq) (answer in V) Numeric Answer:
A chemist designs a galvanic cell that uses these two half-reactions: standard reduction potential half-reaction +...
A chemist designs a galvanic cell that uses these two half-reactions: standard reduction potential half-reaction + O2(9)+4 H (aq)+4e' 2H20) = 1.23 V red Ered Fe+. (аq) Fe3(aq)+e = +0.771 V Answer the following questions about thiss cell Write a balanced equation for the half-reaction that happens at the cathode Write a balanced equation for the half-reaction that happens at the anode Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous...
Please show all steps taken (prefer typed solution) Half-Reaction E ° (V) Ag+ (aq) + e−...
Please show all steps taken (prefer typed solution)
Half-Reaction
E
°
(V)
Ag+ (aq) + e− → Ag (s)
0.7996
Al3+ (aq) + 3e− → Al (s)
−1.676
Au+ (aq) + e− → Au (s)
1.692
Au3+ (aq) + 3e− → Au (s)
1.498
Ba2+ (aq) + 2e− → Ba (s)
−2.912
Br2 (l) + 2e− → 2Br− (aq)
1.066
Ca2+ (aq) + 2e− → Ca (s)
−2.868
Cl2 (g) + 2e− → 2Cl− (aq)
1.35827
Co2+ (aq) + 2e−...
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.26 M )+2e- Red:...
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.26 M )+2e- Red: MnO-4(aq, 1.50 M )+4H+(aq, 2.7 M )+3e-→MnO2(s)+2H2O(l) Compute the cell potential at 25 ∘C.
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.18 M )+2e− Red:...
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.18 M )+2e− Red: MnO−4(aq, 1.65 M )+4H+(aq, 1.9 M )+3e−→ MnO2(s)+2H2O(l) Compute the cell potential at 25 ∘C.
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.29 M )+2e− Red:...
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.29 M )+2e− Red: MnO−4(aq, 1.30 M )+4H+(aq, 1.7 M )+3e−→ MnO2(s)+2H2O(l) show work
12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery...
12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery that... Reduction Half-Reaction F2(g) + 2e →2F(aq) S2082 (aq) + 2e- → 25042 (aq) O2(g) + 4H+ (aq) + 4e → 2H2O(1) Br2(1) + 2e + 2Br (aq) Agt(aq) + e → Ag(s) Fe3+ (aq) + e- → Fe2+ (aq) 126) + 2e → 21 (aq) Cu2+ (aq) + 2e → Cu(s) Sn4+ (aq) + 2e → Sn2+ (aq) S(s) + 2H+ (aq) +...
1. How do I read the half reaction table? 2. If im asked for the best...
1. How do I read the half reaction table?
2. If im asked for the best reducing agent from Cu+, Ag+, F2, and
Fe3+, where do I look first in the table? before the arrow or after
the arrow?
3. Sometimes a value that has originally a positive (V) from the
table it will have the negative sign in a homework problem, and
viceversa, so the question is, how do I use the positive and
negative signs in respect to...
Half-potentials. Homework – Unanswered Due in 8 hours For the reaction: Au(s) + 4H" (aq) + NO3(aq) + 4C"(aq) -- Aucla (aq) + NO(g) + 2H20 the standard potential E=-0.04 V. If the standard potential for the reduction of NO3 aq) to NO(g) is +0.96 V, determine the value of Efor the following half-reaction. AuCl(aq) + 3e - Au(s) + 4C1 (aq) (answer in V) Numeric Answer:
A chemist designs a galvanic cell that uses these two half-reactions: standard reduction potential half-reaction + O2(9)+4 H (aq)+4e' 2H20) = 1.23 V red Ered Fe+. (аq) Fe3(aq)+e = +0.771 V Answer the following questions about thiss cell Write a balanced equation for the half-reaction that happens at the cathode Write a balanced equation for the half-reaction that happens at the anode Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous...
Please show all steps taken (prefer typed solution)
Half-Reaction
E
°
(V)
Ag+ (aq) + e− → Ag (s)
0.7996
Al3+ (aq) + 3e− → Al (s)
−1.676
Au+ (aq) + e− → Au (s)
1.692
Au3+ (aq) + 3e− → Au (s)
1.498
Ba2+ (aq) + 2e− → Ba (s)
−2.912
Br2 (l) + 2e− → 2Br− (aq)
1.066
Ca2+ (aq) + 2e− → Ca (s)
−2.868
Cl2 (g) + 2e− → 2Cl− (aq)
1.35827
Co2+ (aq) + 2e−...
12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery that... Reduction Half-Reaction F2(g) + 2e →2F(aq) S2082 (aq) + 2e- → 25042 (aq) O2(g) + 4H+ (aq) + 4e → 2H2O(1) Br2(1) + 2e + 2Br (aq) Agt(aq) + e → Ag(s) Fe3+ (aq) + e- → Fe2+ (aq) 126) + 2e → 21 (aq) Cu2+ (aq) + 2e → Cu(s) Sn4+ (aq) + 2e → Sn2+ (aq) S(s) + 2H+ (aq) +...
1. How do I read the half reaction table?
2. If im asked for the best reducing agent from Cu+, Ag+, F2, and
Fe3+, where do I look first in the table? before the arrow or after
the arrow?
3. Sometimes a value that has originally a positive (V) from the
table it will have the negative sign in a homework problem, and
viceversa, so the question is, how do I use the positive and
negative signs in respect to...
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