Given the following reduction half-reactions:
Fe3+(aq)+e−→Fe2+(aq)
E∘red=+0.77V
S2O2−6(aq)+4H+(aq)+2e−→2H2SO3(aq)
E∘red=+0.60V
N2O(g)+2H+(aq)+2e−→N2(g)+H2O(l)
E∘red=−1.77V
VO+2(aq)+2H+(aq)+e−→VO2+(aq)+H2O(l)
E∘red=+1.00V
Write balanced chemical equation for the oxidation of Fe2+(aq) by S2O2−6(aq).
Calculate ΔG∘ for this reaction at 298 K.
Calculate the equilibrium constant Kfor this reaction at 298 K.
Write balanced chemical equation for the oxidation of Fe2+(aq) by N2O(g).
Calculate ΔG∘ for this reaction at 298 K.
Calculate the equilibrium constant Kfor this reaction at 298 K.
Write balanced chemical equation for the oxidation of Fe2+(aq) by VO+2(aq).
Calculate ΔG∘ for this reaction at 298 K.
Calculate the equilibrium constant Kfor this reaction at 298 K.
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Given the following reduction half-reactions: Fe3+(aq)+e−→Fe2+(aq) E∘red=+0.77V S2O2−6(aq)+4H+(aq)+2e−→2H2SO3(aq) E∘red=+0.60V N2O(g)+2H+(aq)+2e−→N2(g)+H2O(l) E∘red=−1.77V VO+2(aq)+2H+(aq)+e−→VO2+(aq)+H2O(l) E∘red=+1.00V Write balanced chemica
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
com/mv Problem 20.51 K3 of3 Part G Write balanced chemical equation for the oxidation of Fe (a?) by VO (aq) Fe (ag)eFe (aq) Ered +0.77v S2O% (ag) + 4H (aq) +2e -2H,SO (aq) N20(9) +2H (a)+2eN2(9) H20(U) Part H Calculate AG for this reaction at 298 K k.J 45 PM 5/B/2018 O Type here to search
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 Part A Part complete An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.017M)+2H+(aq,1.3M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l) Calculate the cell potential under these nonstandard concentrations.
NO−3(aq)+4H+(aq)+3e−→NO(g)+2H2O(l) E∘=0.96V ClO2(g)+e−→ClO−2(aq) E∘=0.95V Cu2+(aq)+2e−→Cu(s) E∘=0.34V 2H+(aq)+2e−→H2(g) E∘=0.00V Pb2+(aq)+2e−→Pb(s) E∘=−0.13V Fe2+(aq)+2e−→Fe(s) E∘=−0.45V You may want to reference (Pages 898 - 902) Section 19.4 while completing this problem. Part A Use data from the table above to calculate E∘cell for the reaction. Fe(s)+2H+(aq)→Fe2+(aq)+H2(g) Express your answer using two decimal places.
13. Find the standard reduction potentials for the following half-reactions: So+4H" + 2e-> HSO H2O and Ag" + e- > Ag a) Write the balanced overall reaction for a successful cell made from these two couples. b) Write the line notation for the cell. c) What is Eo for the cell? d) What is the equilibrium constant for the cell reaction at 250C? e) Calculate the ratio of activities of produces and reactants, Q, that will produce a cell voltage...
Consider the following half-reactions: Half-reaction E° (V) 12(s) + 2e - →21"(aq) 0.535V 2H+ (aq) + 2e - → H2(g) 0.000V Cr3+(aq) + 3e —— Cr(s) -0.740V The strongest oxidizing agent is: enter formula The weakest oxidizing agent is: The weakest reducing agent is: The strongest reducing agent is: Will 12(s) reduce Cr3+(aq) to Cr(s)? — Which species can be reduced by H2(g)? If none, leave box blank. Use the References to access important values if needed for this question....
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) 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...
Parts a-j of this question refer to the following cell at 298 K. Complete the diagram of the following voltaic cell: Zinc and iron electrodes, aqueous zinc nitrate, aqueous iron (III) nitrate, potassium nitrate salt bridge. Fe3+ (aq) + 3e- → Fe (s) E°= -0.040 V Ag+(aq) + e- → Ag(s) E°= 0.795 V a. Identify the electrodes as Anode or Cathode. Write the correct metal for each assuming the reactions are spontaneous b. Show the direction of the electron...
Write balanced half-reactions for the following redox reaction: 4Zn+2(aq)+AsH3(g)+8OH−(aq)→ 4Zn(s)+H3AsO4(aq)+4H2O(l) clearly state oxidation reduction