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EDAY THURSDAY mead SATURDAY 27. Calculate for the electrochemical cell below, Pb(s) PbCl(s) CH(aq. 1.0 M)...
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
calculate ecell for the following electrochemical cell at 25 C. Pt(s)| H2 (g,1.00 atm) | H+ (aq, 1.00 M) || (Pb2+ (aq, 0.150 M) | Pb (s) Pb^2+(aq) + 2e- ---> Pb (s) Eo=-0.126 V 2H^+ (aq) + 2e- ---> H2 (g) Eo=0.00 V Thank you!
Using the Nernst equation calculate the cell voltage for: Fe(s) + Cd2+(aq) → Fe2+(aq) + Cd(s) when the [Fe2+] = 0.20 M and [Cd2+] = 1.5 M. Potentially useful information: Fe2+ + 2e− → Fe(s); ε0 = -0.44 V Cd2+ + 2e− → Cd(s); ε0 = -0.40 V
Pb2+(aq) + 2e− ⇌ Pb(s) E° = -0.126 V 2H+(aq) + 2e− ⇌ H2(g) E° = 0.000 V E°cell (in V)= 0.126 V 2. The electrochemical cell is comprised of a Pb electrode in a 1.67 × 100 M solution of Pb2+ (aq) coupled to a Pt electrode in a solution containing H+ (aq) where the pH of the solution is 0.37 and the partial pressure of H2(g) is 0.571 atm. The temperature of the cell is held constant at...
Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2: Fe2+(aq)/Fe(s) and Zn2+(aq)/Zn(s) Which of the following is correct for the working cells? Standard reduction potentials, 298 K, Aqueous Solution (pH = 0): Cl2(g) + 2e --> 2C1-(aq); E° = +1.36 V Fe3+(aq) + e --> Fe2+(aq); E° = +0.77 V Cu2+(aq) + 2e --> Cu(s); E° = +0.34 V 2H+(aq) + 2e --> H2(g); E° = 0.00 V Pb2+(aq) + 2e --> Pb(s);...
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(aq, 0.500 M) || Cu2+(aq, 1.00 M) Cu(s) Candidate 3: Mg(s) | Mg2+(aq, 0.500 M) | Pb2+(aq, 1.00 M)| Pb(s) (a) 6 pts) Choose one of the candidate voltaic cells #1, #2, or #3. Draw a schematic cell diagram for the candidate voltaic cell of choice. Clearly label anode, cathode, electrodes, ions and their concentrations, salt bridge, and the flow of electrons. (b) (5 pts)...
Calculate Ecell for the following electrochemical cell at 25 degreeC Pt(s) | H2(g, 1.00 atm) | H+ (aq, 1.00 M) || Pb2+(aq, 0.150 M) | Pb(s) given the following standard reduction potentials. Pb2+(aq) + 2 e– --> Pb(s) Edegree = –0.126 V 2 H+ (aq) + 2 e– --> H2(g) Edegree = 0.000 V (Please show the steps)
For the following electrochemical cell: 2 Al(s) + 3 Mn?"(aq) 2 Al3+ (aq) + 3 Mn(s) E = 0.48 V what is the value of E (at 298 K) when [AP*] = 1.0 M and [ Mn2'] = 0.050 M? 1.38 V 0.44 V 0.58 V 0.48 V O 0.22 V
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.
What is the standard emf of a galvanic cell made of a Co electrode in a 1.0 M Co(NO32 solution and a Al electrode in a 1.0 M AI(NO3)3 solution at 25°C? 0 cell Standard Reduction Potentials at 25°C Half-Reaction E(V +2.87 +2.07 +1.82 O,(g) 2H (aq)2e0(g)+HO Co3+(aq) + e-_? Co2+(aq) H,02(aq) + 2H"(aq) + 2e-_ 2H20 Cu2+(aq) + 2e-? Cu(s) AgCIs) + Ag(s) + CI(a) S02-(aq) + 4H'(aq) + 2e S02(g) + 2H20 Cu2+(aq) + e-_ Cu+(aq) Sn (aq)...