Overall Balanced reaction :
3 Pb (s) + 2 MnO4-(aq) + 8H+ (aq) ------> 3 Pb2+(aq) + 2 MnO2 (s) + 4H2O (l)
E0cell = E0cathode - E0anode = + 1.81 V
Ecell = E0cell - (0.0592 V /6) log { [Pb2+]3 [MnO2 ]2 [H2O]4 / [Pb]3 [MnO4-]2 [H+]8 } at 25 C
for solids and liquids , activity ~ 1.
Ecell = 1.81 V - (0.0592 V /6) log { [Pb2+]3 / [MnO4-]2 [H+]8 } = 1.81 V - (0.0592 V /6) log { [0.21]3 / [1.4]2 [1.5]8 }
Ecell = 1.81 V - ( -0.22 )
Ecell = 2.03 V
An electrochemical cell is based on these two half-reactions: Ox: Pb (8) Pb²+ (aq, 0.21 mol...
An electrochemical cell is based on these two half-reactions: Ox: Pb(s) Pb?" (aq, 0.21 mol L ')+2 e Erode = -0.13 (V) Red: Mno. (aq, 1.40 mol L-)+4 H+ (aq, 1.5 mol L-') +3 e MnO2 (s) + 2 H2O (1) Ecathode = 1.68 (V) Part A Compute the cell potential at 25°C. Express your answer to two decimal places and include the appropriate units. Ecell = Value Units Submit Request Answer
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.26 M )+2e- Red: MnO-4(aq, 1.50 M )+4H+(aq, 2.7 M )+3e-→MnO2(s)+2H2O(l) Compute the cell potential at 25 ∘C.
יי Review 1 Constants l Periodic Table An electrochemical cell is based on these two half-reactions Ox: Sn(s) → Sn2+ (aq, 1.80 mol L i) + 2 e- Eanode-_0.14 (V) Red. CO2 (g, 0.110 bar) + e-→ Cio, (aq, 1.55 mol L i) E cathode = 0.95 (V) Part A Compute the cell potential at 25 C Express your answer to two decimal places and include the appropriate units
19. Calculate EceLL (in V to two decimal places) for an electrochemical cell based on the following half-reactions at equilibrium. In addition, determine ΔG° (in kJ mol-1 to two decimal places) for the reaction under standard conditions (i.e. all concentrations are 1.0 M) and predict the magnitude of K (e.g. very small, very large, etc.) E (o)0.34 V; E red) 0 9 = 1.68 V oxidation : Cu (s) → Cu2+ (aq, .010 M) + 2e- Reductin: MnO4 (aq, 2.0...
A voltaic cell is based on two lead half cells: Pb+2(aq) + 2 e- → Pb(s) Eo = -0.13 v One half-cell has [Pb(NO3)2] = 0.25 M and one half-cell has [Pb(NO3)2] = 0.0020 M What is the initial potential produced by the voltaic cell. Express your answer in units of volts with 2 significant digits, but do not include the units on the submitted answer.
Calculate e cell for the electrochemical cell below, Pb(s) |Pb2+(aq, 1.0 M) || Fe2+(aq, 1.0 M) | Fe(s) given the following reduction half-reactions. Pb2+(aq) + 2 e– ® Pb(s) E° = –0.126 V Fe2+(aq) + e– ® Fe(s) E° = –0.44 V
A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential Cl 2 (g) + 2 e − → 2 Cl − (aq) = E 0 red + 1.359 V MnO − 4 (aq) + 2 H 2 O (l) + 3 e − → MnO 2 (s) + 4 OH − (aq) = E 0 red + 0.59 V cathode half reaction anode half reaction overall reaction cell potential at standard state
An electrochemical cell is based on the following two half-reactions: Part A oxidation: Sn (s) +Sn2+ (aq, 1.70 M)+2e reduction: CIO2(g, 0.265 atm )+e+C10(aq, 2.00 M) You may want to reference (Pages 865 - 869) Section 19.6 while completing this problem. Compute the cell potential at 25°C. Express the cell potential to three significant figures. ME PO ΑΣΦ ? Ecco V Submit Request Answer
Design a voltaic cell with the following two reduction half-reactions: Ag+(aq) + e− ⟶ Ag(s) Eo = 0.80 V Pb2+(aq) + 2 e− ⟶ Pb(s) Eo = −0.13 V Calculate Eocell and the equilibrium constant K for the voltaic cell at 298 K. Click here for a copy of Final Exam cover sheet.
The standard half-cell reactions of lead acid battery is: 2) ЕФN Half-cell reduction reaction form: red PbOo2.(s) 4H+ + so2 4, (aq) PbSO4,()+2H20 Red 2e 1.69 Рo) + So?- 4.(aq) -Ox -0.36 Pbs04.(s)2e Pb()PbO2,(s)+ 2H2SO4,(ag) 2P6SO4.(6) +2H20D 2.05 Tot. a) Determine the full cell reaction of lead acid batter and the total cell potential when discharging the battery b) The half-cell potentials are given according to a reference potential. Describe this reference potential. What is the point of a reference...