Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2 ] = 0.774 M and [Sn2 ] = 0.0190 M. Standard reduction potentials can be found here.
| Reduction Half-Reaction | Standard Potential Ered° (V) |
| F2(g) + 2e– → 2F–(aq) | +2.87 |
| O3(g) + 2H3O+(aq) + 2e– → O2(g) + 3H2O(l) | +2.076 |
| Co3+(aq) + e– → Co2+(aq) | +1.92 |
| H2O2(aq) + 2H3O+(aq) + 2e– → 2H2O(l) | +1.776 |
| N2O(g) + 2H3O+(aq) + 2e– → N2(g) + 3H2O(l) | +1.766 |
| Ce4+(aq) + e– → Ce3+(aq) | +1.72 |
| PbO2(s) + SO42–(aq) + 4H3O+(aq) + 2e– → PbSO4(s) + 6H2O(l) | +1.6913 |
| MnO4–(aq) + 4H3O+(aq) + 3e– → MnO2(s) + 6H2O(l) | +1.679 |
| NiO2(s) + 4H3O+(aq) + 2e– → Ni2+(aq) + 6H2O(l) | +1.678 |
| 2HClO(aq) + 2H3O+(aq) + 2e– → Cl2(g) +2H2O(l) | +1.611 |
| H5IO6(s) + H3O+(aq) + 2e– → IO3–(aq) + 4H2O(l) | +1.601 |
| 2NO(g) + 2H3O+(aq) + 2e– → N2O(g) + 3H2O(l) | +1.591 |
| MnO4–(aq) + 8H3O+(aq) + 5e– → Mn2+(aq) + 12H2O(l) | +1.507 |
| Au3+(aq) + 3e– → Au(s) | +1.498 |
| PbO2(s) + 4H3O+(aq)+ 2e– → Pb2+(aq) + 6H2O(l) | +1.455 |
| ClO3–(aq) + 6H3O+(aq) + 6e– → Cl–(aq) + 9H2O(l) | +1.451 |
| BrO3–(aq) + 6H3O+(aq) + 5e– → 1/2Br2(l) + 9H2O(l) | +1.482 |
| 2ClO4–(aq) + 16H3O+(aq) + 14e– → Cl2(g) + 24H2O(l) | +1.39 |
| ClO4–(aq) + 8H3O+(aq) + 8e– → Cl–(aq) + 12H2O(l) | +1.389 |
| Cl2(g) + 2e– → 2Cl–(aq) | +1.36 |
| ClO4–(aq) + 6H3O+(aq) + 6e– → ClO–(aq) + 9H2O(l) | +1.36 |
| HBrO(aq) + H3O+(aq) + 2e– → Br– + 2H2O(l) | +1.331 |
| Cr2O72–(aq) + 14H3O+(aq) + 6e– → 2Cr3+(aq) + 21H2O(l) | +1.232 |
| O2(g) + 4H+(aq) + 4e– → 2H2O(l) | +1.23 |
| MnO2(s) + 4H3O+(aq) + 2e– → Mn2+(aq) + 6H2O(l) | +1.224 |
| 2IO3–(aq) + 12H3O+(aq) + 10e– → I2(s) + 18H2O(l) | +1.195 |
| ClO4–(aq) + 2H3O+(aq) + 2e– → ClO3–(aq) + 3H2O(l) | +1.189 |
| Pt2+(aq) + 2e– → Pt(s) | +1.18 |
| Br2(aq) + 2e– → 2Br–(aq) | +1.0873 |
| Br2(l) + 2e– → 2Br–(aq) | +1.07 |
| NO2(g) + 2H3O+(aq) + 2e– → NO(g) + 3H2O(l) | +1.03 |
| NO3–(aq) + 4H3O+(aq) +3e– → NO(g) + 6H2O(l) | +0.957 |
| 2Hg2+(aq) + 2e– → Hg22+(aq) | +0.920 |
| Hg2+(aq) + 2e– → Hg(l) | +0.851 |
| ClO–(aq) + H2O(l) + 2e– → Cl–(aq) + 2OH–(aq) | +0.81 |
| Ag+(aq) + e– → Ag(s) | +0.80 |
| Hg22+(aq) + 2e– → 2Hg(l) | +0.7973 |
| Fe3+(aq) + e– → Fe2+(aq) | +0.771 |
| Ni(OH)2(s) + 2e– → Ni(s) + 2OH–(aq) | +0.72 |
| p-benzoquinone + H3O+(aq) + 2e– → hydroquinone + H2O(l) | +0.6992 |
| O2(g) + 2H3O+(aq) + 2e– → H2O2(l) + 2H2O(l) | +0.695 |
| MnO4–(aq) + 2H2O(l) + 3e– → MnO2(s) + 4OH–(aq) | +0.595 |
| I2(s) + 2e– → 2I–(aq) | +0.54 |
| I3–(aq) + 2e– → 3I–(aq) | +0.536 |
| Cu+(aq) + e– → Cu(s) | +0.52 |
| O2(g) + 2H2O + 4e– → 4OH–(aq) | +0.401 |
| Fe(CN)63–(aq) + e– → Fe(CN)64–(aq) | +0.358 |
| Cu2+(aq) + 2e– → Cu(s) | +0.34 |
| Hg2Cl2(s) + 2e– → 2Hg(l) + 2Cl–(aq) | +0.26808 |
| HAsO2(s) + 3H3O+(aq) + 3e– → As(s) + 5H2O | +0.248 |
| AgCl(s) + e– → Ag(s) + Cl-(aq) | +0.22233 |
| Cu2+(aq) + e– → Cu+(aq) | +0.153 |
| Sn4+(aq) +2e– → Sn2+(aq) | +0.151 |
| S(s) + 2H3O+(aq) + 2e– → H2S(s) + 2H2O(l) | +0.14 |
| NO3–(aq) +2H2O(l) + 3e– → NO(g) + 4OH–(aq) | +0.109 |
| N2(g) + 8H3O+(aq) + 6e– → 2NH4+(aq) +8H2O(l) | +0.092 |
| S4O62–(aq) + 2e– → 2S3O32–(aq) | +0.08 |
| AgBr(s) + e– → Ag(s) + Br–(aq) | +0.07133 |
| 2H+(aq) + 2e– → H2(g) | 0.00 |
| Fe3+(aq) + 3e– → Fe(s) | -0.04 |
| [Co(NH3)6]3+(aq) + e– → [Co(NH3)6]2+(aq) | -0.108 |
| Pb2+(aq) + 2e– → Pb(s) | –0.13 |
| Sn2+(aq) + 2e– → Sn(s) | –0.14 |
| O2(g) + 2H2O(l) + 2e– → H2O2(l) + 2OH–(aq) | –0.146 |
| AgI(s) + e– → Ag(s) + I– (aq) | –0.15224 |
| CO2(g) + 2H3O+(aq) + 2e– → HCO2H(s) + 2H2O(l) | –0.199 |
| Cu(OH)2(s) + 2e– → Cu(s) + 2OH–(aq) | –0.222 |
| Ni2+(aq) + 2e– → Ni(s) | –0.26 |
| Co2+(aq) + 2e– → Co(s) | –0.28 |
| PbSO4(s) + 2e– → Pb(s) + SO42–(aq) | –0.3588 |
| SeO32–(aq) + 3H2O(l) + 4e– → Se + 6OH–(aq) | –0.366 |
| Cd2+(aq) + 2e– → Cd(s) | –0.403 |
| Cr3+(aq) + e– → Cr2+(aq) | –0.407 |
| Fe2+(aq) + 2e– → Fe(s) | –0.44 |
| NO2–(g) + H2O(l) + 3e– → NO(g) + 2OH–(aq) | –0.46 |
| S(s) + 2e– → S2–(aq) | –0.48 |
| 2CO2(g) + 2H3O+(aq) + 2e– → H2C2O4(s) + H2O(l) | –0.49 |
| TiO2(s) + 4H3O+ + 2e– → Ti2+(aq) + 6H2O(l) | –0.502 |
| Au(CN)2–(aq) + e– → Au(s) + 2CN–(aq) | –0.60 |
| Cr3+(aq) + 3e– → Cr(s) | –0.74 |
| Zn2+(aq) + 2e– → Zn(s) | –0.76 |
| Cd(OH)2(s) + 2e– → Cd(s) + 2OH–(aq) | –0.809 |
| 2H2O(l) + 2e– → H2(g) + 2OH–(aq) | –0.83 |
| Ti3+(aq) + e– → Ti2+(aq) | –0.85 |
| H3BO3(s) + 3H3O+ + 3e– → B(s) + 6H2O(l) | –0.8698 |
| Cr2+(aq) + 2e– → Cr(s) | –0.91 |
| SO42–(aq) + H2O(l) + 2e– → SO32–(aq) + 2OH–(aq) | –0.93 |
| CNO–(aq) + H2O(l) + 2e– → CN–(aq) + 2OH–(aq) | –0.970 |
| [Zn(NH3)4]2+(aq) + 2e– → Zn(s) + 4NH3(aq) | –1.04 |
| Mn2+(aq) + 2e– → Mn(s) | –1.185 |
| Cr(OH)3(s) + 3e– → Cr(s) + 3OH–(aq) | –1.48 |
| Ti2+(aq) + 2e– → Ti(s) | –1.630 |
| Al3+(aq) + 3e– → Al(s) | –1.66 |
| Al(OH)3(s) + 3e– → Al(s)+3OH–(aq) | –2.31 |
| Mg2+(aq) + 2e– → Mg(s) | –2.38 |
| Mg(OH)2(s) + 2e– → Mg(s) + 2OH–(aq) | –2.69 |
| Na+(aq) + e– → Na(s) | –2.71 |
| Ca2+(aq) + 2e– → Ca(s) | –2.87 |
| Ba2+(aq) + 2e– → Ba(s) | –2.912 |
| K+(aq) + e– → K(s) | –2.931 |
| Ba(OH)2(s) + 2e– → Ba(s) + 2OH–(aq) | –2.99 |
| Ca(OH)2(s) + 2e– → Ca(s) + 2OH–(aq) | –3.02 |
| Cs+(aq) + e– → Cs(s) | –3.026 |
| Li+(aq) + e– → Li(s) | –3.04 |
Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2...
Using standard reduction potential in aqueous solutions at 25c Table, which substance is most likely to be oxidised by O2 (g) in acidic aqueous solution? Select one: a. Br2 (l) b. Br- (aq) c. Ni2+ (aq) d. Ag (s) e. Cu2+ (aq) Cathode (Reduction) Half-Reaction Standard Potential E° (volts) Li+(aq) + e- -> Li(s) -3.04 K+(aq) + e- -> K(s) -2.92 Ca2+(aq) + 2e- -> Ca(s) -2.76 Na+(aq) + e- -> Na(s) -2.71 Mg2+(aq) + 2e- -> Mg(s) -2.38 Al3+(aq)...
croHissit Song "14. a) Calculate the standard emf and write the overall equation for the cell described as: Croaq) Haq) + (aq) → Craq) + 2/8) + H2O(D Cr₂O7991+ 4H₂011 +36-7 reducing Oxidizing 2 croren + 4H20 (0) +36 --> CrotsstSoH + 5% 3(21 691 -> Izintze-) 0.406 14 7 .0 I BOV overall eqn: 200 2 06112 --> 2010 He(s) tel +3120) emf: 0.106 b) Calculate the emf obtained by this cell (based on part a above) from the...
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−...
Calculate the equilibrium constant for each of the reactions at 25∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction E∘(V) Fe3+(aq)+3e− →Fe(s) -0.036 Sn2+(aq)+2e− →Sn(s) -0.14 Cu2+(aq)+2e− →Cu(s) 0.16 O2(g)+2H2O(l)+4e− →4OH−(aq) 0.40 Cl2(g)+2e− →2Cl− 1.36 I2(s)+2e− →2I− 0.54 Part B O2(g)+2H2O(l)+2Cu(s)→4OH−(aq)+2Cu2+(aq) (Express your answer using two significant figures. ) Part C Cl2(g)+2I−(aq)→2Cl−(aq)+I2(s) (Express your answer using two significant figures.)
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If you can explain how to solve for each part please.
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Express your answer using two significant figures.
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Express your answer using two significant figures.
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Express your answer using two significant figure.
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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...
Calculate the equilibrium constant for each of the reactions at 25∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction E∘(V) Fe3+(aq)+3e− →Fe(s) -0.036 Sn2+(aq)+2e− →Sn(s) -0.14 Ni2+(aq)+2e− →Ni(s) -0.23 O2(g)+2H2O(l)+4e− →4OH−(aq) 0.40 Br2(l)+2e− →2Br− 1.09 I2(s)+2e− →2I− 0.54 A) 2Fe3+(aq)+3Sn(s)→2Fe(s)+3Sn2+(aq) (answers are not 4.1x10^5, 3.3x10^3, 2.7x10^10, or 2.6x10^10) B) O2(g)+2H2O(l)+2Ni(s)→4OH−(aq)+2Ni2+(aq) C) Br2(l)+2I−(aq)→2Br−(aq)+I2(s) (answer is not 1.7x10^18)