4) (14 pt) Calculate the Eº for the reaction between S2- and Fe3+ that yields SO42-...
For a particular redox reaction, SO32- is oxidized to SO42- and Fe3+ is reduced to Fe2+. Complete and balance the equation for this reaction in basic solution. The phases are optional. balanced reaction: _______
3. For the cell represented by Pt|Fe3+(0.25 M), Fe2+(0.025 M)||Ce4+(0.035 M), Ce3+(0.050 M)|Pt E o (Fe3+/Fe2+ = +0.771V, Eo Ce4+/Ce3+ = + 1.44V a) Draw the complete cell represented by the above line diagram and label all the components. b) Write the half-cell reactions and complete redox reaction of the cell. c) Calculate the standard cell potential. d) Calculate the cell potential at the non-standard conditions provided above. e) Was the cell reaction spontaneous?
7.50 Consider the following cell. Pt Cu2+, Cu+ Fe2+, Fe3+ | Pt (a) What is the cell reaction? (b) What is the standard electro- motive force of the cell at 298.15 K? (c) Calculate Ar Gº for the cell reaction from the standard electromotive force. (d) Calcu- late ArGº for the cell reaction using the AfGº values for the ions in Table C.2. (e) Calculate ArGº for the cell reaction using the AfHº values and Sº values in Table C.2.
please answer in 10m
5. Balance each reaction, calculate the net Eº, and declare each reaction as being spontaneous or non- spontaneous. (9) a) Sn?' (aq) + Cr(s) → Sn(s) + Cr?" (aq) net Eº = b) Ag+ (aq) + Mn(s) Ag(s) + Mn2(aq) net Eº = c) Fe2(aq) + Cu(s) → Fe(s) + Cu?"(aq) net Eº =
Calculate the equilibrium constant under normal conditions of the reaction: Fe3+ + Cu+ ⇄ Fe2+ + Cu2+ DATA: Fe3+ / Fe2+ = 0.714V; Cu2+ / Cu+ = 0.158V
Consider the following cell diagram: Pt(s) | Fe3+(aq) , Fe2+(aq) || Cl–(aq) | Cl2(g) | Pt(s) The reaction utilized by this cell is Question 8 options: Fe2+(aq) + 2Cl–(aq) --> Fe(s) + Cl2(g) Fe(s) + Cl2(g) --> Fe2+(aq) + 2Cl–(aq) 2Fe3+(aq) + 2Cl–(aq) --> 2Fe2+(aq) + Cl2(g) Fe3+(aq) + Cl–(aq) --> Fe2+(aq) + 1/2Cl2(g) 2Fe2+(aq) + Cl2(g) --> 2Fe3+(aq) + 2Cl–(aq)
Consider the following cell: Pt(s) | Fe3+ (aq). Fe2(aq) | CIF (aq) C12(e) Pt(s) If the standard reduction potentials of the Fe3+/Fe2+ and Cl2/Cl" couples are +0.77 and +1.36 V, respectively, calculate the value of Efor the given cell. +1.00 V O +1.77 v +0.59 V +2.13 V +0.95 V
Use standard reduction potentials to calculate ΔG° for the reaction: S2- + 4 Cl2 + 12 H2O →SO42- + 8 Cl- + 8 H3O+ E°SO42-/S2-, H3O+ = +0.149, E°Cl2/Cl- = +1.35827 answer in kj
4. a) Determine Eº for the cell Pt (s) | Cl2 (g) | Cl− (aq) || Pb2+ (aq), H+ (aq) | PbO2 (s). b) What is the Standard Gibbs free energy of this reaction?
Calculate E°(cell) for the reaction, 2 103"(aq) + 10 Fe2+(aq) <=> 10 Fe3* (aq) + typen the reductor per Fe3+ (aq) + e* <=> Fe2+(aq), E = 0.87 V 2 103(aq) + 10 e<=> 12(aq), E° = 1.10 V A. -7.60 V B. 0.23 V C. -1.97 v D. 1.97 v E. -0.23 V