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 complete the sentences.
Words: HCL Above Below Positive Negative HNO3 |
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)...
Table 20.1 Half Reaction E°(V). F2 (g) + 2e →2F (aq) +2.87 Cl2 (g) + 2e → 2CV (aq) +1.359 Br2 (1) + 2e → 2Br (aq) +1.065 O2 (g) + 4H+ (aq) + 4e → 2H20 (1)+1.23 Agt te → Ag (s) +0.799 Fe3+ (aq) + € → Fe2+ (aq) +0.771 12 (s) + 2e → 21+ (aq) +0.536 Cu2+ + 2e → Cu(s) +0.34 2H+ + 2e → H2 (g) Pb2+ + 2e → Pb (s) -0.126 Ni2+...
Classify the half‑reactions as reduction half‑reactions or oxidation half‑reactions. H2(g)⟶2H+(aq)+2e−H2(g)⟶2H+(aq)+2e− 12O2(g)+2H+(aq)+2e−⟶H2O(g)12O2(g)+2H+(aq)+2e−⟶H2O(g) Cd(s)+2OH−(aq)⟶Cd(OH)2(s)+2e−Cd(s)+2OH−(aq)⟶Cd(OH)2(s)+2e− 2NiO(OH)(s)+2H2O(l)+2e−⟶2Ni(OH)2(s)+2OH−(aq)2NiO(OH)(s)+2H2O(l)+2e−⟶2Ni(OH)2(s)+2OH−(aq) Fe(s)⟶Fe2+(aq)+2e−Fe(s)⟶Fe2+(aq)+2e− oxidation reduction reduction oxidation reduction
Calculate the equilibrium constant for each of the reactions at 25 ∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction E∘(V) Pb2+(aq)+2e− →Pb(s) -0.13 Zn2+(aq)+2e− →Zn(s) -0.76 Br2(l)+2e− →2Br−(aq) 1.09 Cl2(g)+2e− →2Cl−(aq) 1.36 MnO2(s)+4H+(aq)+2e− →Mn2+(aq)+2H2O(l) 1.21 Pb2+(aq)+2e− →Pb(s) -0.13 Br2(l)+2Cl−(aq)→2Br−(aq)+Cl2(g) Express your answer using two significant figures.
Refer to this table of reduction potentials to answer the questions. Reduction half-reaction Potential (V) +2.87 Electrolytic cells use electricity to cause a nonspontaneous redox reaction to occur. An electrolytic cell is constructed using the following components: • a power source, such as a battery, • the substance that will undergo electrolysis, and • two inert electrodes (usually platinum), which serve as the electrical connection between the power source and the substance undergoing electrolysis. As with any cell, oxidation occurs...
A. Half-reaction E° (V) Cl2(g) + 2e-2Cl-(aq) 1.360V Sn2+(aq) + 2e-Sn(s) -0.140V Cr3+(aq) + 3e-Cr(s) -0.740V (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will Cl2(g) oxidize Cr(s) to Cr3+(aq)? (6) Which species can be oxidized by Sn2+(aq)? If none, leave box blank. B. Half-reaction E° (V) Cl2(g) + 2e-2Cl-(aq) 1.360V Cu2+(aq) + 2e-Cu(s) 0.337V Mn2+(aq) + 2e-Mn(s) -1.180V (1) The...
Consider the following half-reactions: Half-reaction E° (V) F2(g) +2e - →2F (aq) 2.870V 2H*(aq) + 2e - H2(g) 0.000V Cr3+ (aq) + 3e — Cr(s) -0.740V (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will F2(g) oxidize Cr(s) to Cr3+ (aq)? (6) Which species can be oxidized by H(aq)? If none, leave box blank.
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) +...
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−...
Consider the following half-reactions and their reduction potentials (volt Au (aq) +3e- Au(s) +1.50 petad) +2 e' → Pt(s) +1.20 Co2+(aq) + 2 e' → Cols) -0.28 Mn? (aq)+2e Mn/s) -1.18 Which one of the following statements is correct? O A Ptis) can reduce Ht in aqueous solution OB. P (ag) can be reduced by Cols) OC. A (aa) is the weakest oxidizing agent D. Mn? (aq) can oxidize Au (s) OE. Auls) is the strongest reducing agent ous Save...
Standard Reduction Please write your answers here Reduction Half-Reactin Potential (V) F2(g) + 2e-→ 2F-(aq) S2082 (ag) +2e-2SO42(ag) O2(g) + 4H(a)+ 4e 2H200) +2.87 +2.01 +1.23 +1.09 +0.80 +0.77 +0.54 +0.34 +0.15 +0.14 0.00 0.14 0.26 0.44 0.74 0.76 0.83 1.18 2.71 3.04 2 4 Ag+(aq) + e-→ Ag(s) Fe3+(ag)eFe2*(aq) 20)+ 2e- 21(aq) Cu2(ag)+ 2e Cus) SAMPLE QUIZ 4 S(s) + 2H+(aq) + 2e. → H2S(g) 2H(a)+ 2eH2g) Sn2(ag) 2e Sng) 1. What is the purpose of the salt bridge...