An electrochemical cell is based on these two half-reactions: Ox: Pb(s) Pb?" (aq, 0.21 mol L...
An electrochemical cell is based on these two half-reactions: Ox: Pb (8) Pb²+ (aq, 0.21 mol L-')+2 e Enode -0.13 (V) Red: MnO, (aq, 1.40 mol L-1) + 4H+ (aq, 1.5 mol L-) + 3 e MnO, (s) + 2 H20 (1) Eestbode - 1.68 (V) Part A Compute the cell potential at 25°C. Express your answer to two decimal places and include the appropriate units. T • Ea ? HẢ • Value • v Ecou -
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.
Find Ecell for an electrochemical cell based on the following reaction with [MnO4−]=1.50M, [H+]=1.50M, and [Ag+]=0.0140M. E∘cell for the reaction is +0.88V. MnO4−(aq)+4H+(aq)+3Ag(s)→MnO2(s)+2H2O(l)+3Ag+(aq) Find for an electrochemical cell based on the following reaction with , , and . for the reaction is . 0.75 V 1.01 V 0.84 V 0.92 V
The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt($)|H2(9, 1.0 bar) H+ (aq, 1.0 mol L-Aus+ (aq,? mol L-)|Au(s). Part A What is the concentration of Au3+ in the solution if Ecell is 1.23 V ? Express your answer using two significant figures. YO ALC * * O O ? Au3+1 = mol L-1 Submit Request Answer
MISSED THIS? Read Section 20.6 (Page) An electrochemical cell is based on the following two half-reactions: Part A oxidation: Sn (s) +Sn2+ (aq, 1.80 M )+2e reduction: C1O2(g, 0.210 atm )+e- +C10] (aq, 1.70 M) Compute the cell potential at 25°C. Express the cell potential to three significant figures. ΤΕΙ ΑΣφ ? Ecell = .839 Submit Previous Answers Request Answer X Incorrect; Try Again; One attempt remaining Provide Feedback
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
Scores A Review | Constants | Periodic Table e Text Standard Reduction Half-Cell Potentials at 25°C Assignments Half-Reaction Course Tools + Pb2+ (aq) + 2 H2O(l)| PbO2(s) + 4 H+ (aq) + 2 e Cl2 (9) +2 e +2 CT (aq) Pb2+ (aq) +2 e e → Pb(s) + Pb(s) E° (V) 1.46 1.36 -0.13 Part B Write a balanced equation for the overall reaction. Express your answer as a chemical equation. Identify all of the phases in your answer....