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Given the standard reduction potentials S2032-(aq) + 60H- 25032-(aq) + 3H2O(1) + 4e + (aq) E°...
Use the following reduction potentials to answer questions 208)+2e 02(g) + 4H(aq) +4e- Br20 +2e 2) +2e Niat) + 2e- 2H20u +2e 4-7 Eoed 1.87V Ered = 1.40V Eoed 1.09v Ered = 0.54 V e0.23v Ered =-0.41 V ed0.73V Ere,--3.04V 2H2O(1) ー→ 21(aq) Ni(s) --+ ー→ (aq) aq → Cr Litaa) e ー→ Li(s)
Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG for the following redox reaction. Round your answer to 3 significant digits Mn (aq)+2H,O()+2Fe (aq)MnO, ()+4H (aq)+2Fe (aq) dhData Cu (aq) + e Cu (5) F2 (0)+2e2F (aq) Fe (aq) +2e Fe (s) Fe (aq) + eFe2 (aq) Fe (aq) + 3e Fe (s) 2.866 X ? -0.447 0.771 -e.037 2H (aq)+2e H (0) e.000 2H)O (I)+2e H2 (a) +20H(aq) -0.8277 1.776 H2O2 (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...
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) +...
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)
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...
A certain half-reaction has a standard reduction potential E = -0.45 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that must provide at least 1.00 V of electrical power. The cell will operate under standard conditions Note for advanced students: assume the engineer requires this half-reaction to happen at the cathode of the cell. 0-0 . " Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell...
Given the following standard reduction potentials: H^+ (aq) + 2e^- rightarrow H_2(g) E degree = 0.00 V Fe^3+ (aq) + 2e^- rightarrow Fe(s) E degree = -0.43 V (a) What is the cell potential by combining the above two half-reactions to make a working voltaic cell (same as galvanic cell)? (b) Which species will be oxidized in anode? Write the half-reaction for the anode. (c) Write the overall reaction and balance the chemical equation for this working voltaic cell. (d)...
#17. Electrochemistry: Consider these two entries from a fictional table of standard reduction potentials. X^2+ (aq) + 2e- --> X(s) E = 2.03 V Y^2+ (aq) + 2e- --> Y(s) E = 0.20 V Which pair of species will react under standard conditions at 25 degrees celsius? (1) X and Y (2) X^2+ and Y^2+ (3) X and Y^2+ (4) X^2+ and Y
The following standard reduction potentials have been determined for the aqueous chemistry of ytterbium: Yb3+(aq) + e- -----> Yb2+(aq) E° = -1.050 V Yb2+(aq) + 2e- -----> Yb(s) E° = -2.760 V Calculate the equilibrium constant (K) for the disproportionation of Yb2+(aq) at 25 °C. 3Yb2+(aq) <----->Yb(s) + 2Yb3+(aq) K =