16.52 Balance each of the following unbalanced equations for cell reactions in acidic conditions, then calculate...
- cellpotentials and standard reduction potential table. -Determining the Nernst equation and finding the Faraday constant. If you can explain how to solve for each part please. 25°C Standard Reduction Potentials in Aqueous solution a 2.87 Reduction half-reaction 2F (a 1.77 2H2O 1.692 2e 2H (a Au (s 1.085 PbSO4 (s) 2H20. Au (ag) 2e 1.51 4H20 Mn 1.50 5e 8H (a Mno4 (a Au(S 1.36 3e 2Cl (aq 1.33 2e 2Cr3 (ag) 7H20 C12 6e- 1.229 14H 2H2O 1.08...
Using the standard reduction potentials listed, calculate the equilibrium constant for each of the following reactions at 298 K. A) Fe(s)+Ni2+(aq)→Fe2+(aq)+Ni(s) Express your answer using two significant figures. B) Co(s)+2H+(aq)→Co2+(aq)+H2(g) Express your answer using two significant figures. C) 10Br−(aq)+2MnO−4(aq)+16H+(aq)→2Mn2+(aq)+8H2O(l)+5Br2(l) Express your answer using two significant figure. E°(V) -0.83 +0.88 +1.78 +0.79 Half-Reaction E°(V) Half-Reaction Ag+ (aq) + - Ag(s) +0.80 2 H20(1) + 2 e — H2(8) + 2 OH+ (aq) AgBr(s) + - Ag(s) + Br" (aq) +0.10 HO2...
Question 5 Using the table of standard reduction potentials shown below, calculate the standard cell potential for a battery based on the following reactions. • O2 + 4H+ + 4e + 2 H20 • 2 Cu + 2 Cu+2 + 4e Reduction Half-Reaction F2 + 2e + 2F MnO, +8 H+ + 5e + Mn+2 +4 H20 Cl2 + 2e → 2C O2 + 4H+ + 4e + 2 H2O Agt! + e + Ag Fet3 + e - Fet2...
Question 4 Using the table of standard reduction potentials shown below, calculate the standard cell potential for a battery based on the following reactions. • MnO4 +8H+ + 5e + Mn+2 +4 H20 . 5 Ag + 5 Ag+1 +5e Reduction Half-Reaction F2 +2e + 2F MnO4 + 8 H+ + 5e + Mn+2 + 4H2O Cl2 + 2e + 20 O2 + 4H+ + 4e + 2 H2O Ag++ e + Ag Fet3 Fe + Fet2 O2 + 2...
this is all we were given in class Calculate the value of AG for the following reaction if [H2O2]i = [Fe] =1.0 M [OH']; = 1.3 x10-M, and [Fe3+] = 0.50 M. In which direction will this reaction have to shift in order to reach equilibrium? Explain your answer. Fe2(aq) + H2O2(aq) 5 Fe3+ (aq) + OH(aq) Standard-State Reduction Potentials and Half Rache Best reducing agents K + eK Bat +2 Ba Ca" + 2e Ca Na + c N...
Question 7 Using the table of standard reduction potentials shown below, calculate the standard cell potential for a battery based on the following reactions. • Pet2 + 2e + Fe . 2 Li + 2 Li + 2e Reduction Half-Reaction F2 +2e + 2F MnO + 8H+ Se + Mn+2+ 4 H 0 Cl2 + 2e + 20 02 + 4H' + 4e + 2 H2O Ag+ e -- Ag Fet) + e + Fe2 O2 + 2 H2O +...
8-10 8. Determine the reduction potential for H2O(l) in neutral water: [H"]= [OH = 10? (6 points) (P. = 1 atm) 9. A cobalt electrode is placed in 1.00 L of I M CO(NO3)2(aq) and a chromium electrode is placed in 1.00 L of IM Cr(NO3)3(aq). Electrodes are connected by a wire through a voltmeter and the solutions are connected by a salt bridge. (16 points) a. The reduction reaction will be: b. The oxidation reaction will be: c. The...
need help for half cell potentials pls calculate step by step (NOTE - Remember that the positive electrode is attached to the red wire and the negative electrode is attached to the black wire.) Electrode Systems Used Anode (oxidation) Negative Cathode (reduction) Positive Measured Potential (V) Positive (Ecu) Copper and silver Cu() - Cu2+ + 2e" Ag+ +1e Ag) 0.432 V Zinc + Silver Zn cs + 2n**+ Zé dat + leº nAg (s) 1.484 V Copper & Zinc Zn(s)...
Write the half reactions and overall reaction for each cell with calculated overall potentials as shown in Table 5-1. (Note: for the iron solutions the Nernst equation must be used) Pb(s) | Pb(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) Cu(s) Zn(s) | Zn(NO3)2(0.1M) || Cu(NO3)2 (0.1M) Cu(s) Cds) | Ca(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) | Cu(s) Cu() Cu(NO3)2(0.1M) Il Fe (0.1M/Fe? (0.1M graphite Pb(s) Pb(NO3)2(0.1M) Il Fe3(aq) (0.1M)/ Fe2(aq) (0.1MI graphite(s) Zns | Zn(NO3)2 (0.1M) || Pb(NO3)2 (0.1M) | Pb(s) Cdis Ca(NO3)2...
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...