only information present For the following electrochemical cell: Cd | Cd2+(aq) || Pb2+(aq) | Pb First,...
X) The answers: a) 0.315 V c) 0.273 V d) 1.80 M 10. For the voltaic cell: Cd(s) Cd2 (0.100 M, 25 mL) || Pb2 (2.000 M, 25 mL) | Pb(s) a. What is Eel initially? (E 0.277 v) b. If the cell is allowed to operate spontaneously, will Eel increase, decrease or remain constant with time? Explain. c. What will be Ee when [Pb2"] has fallen to 0.900 M? [Hint: Both concentrations change. Why?] d. What will be [Cd2"]...
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 ∆G° (kJ/mol) for the following electrochemical cell: Cd(s) | Cd2+(aq) || Ni2+(aq)| Ni(s)
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...
A voltaic cell consists of an Al/Al3+ half-cell and a Cd/Cd2+ half-cell. Calculate {Cd2+} when {Al3+} = 0.306 M and Ecell = 1.27 V. Use reduction potential values of Al3+ = -1.66 V and for Cd2+ = -0.40 V.
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
Calculate ΔG° for the electrochemical cell Pb(s) | Pb2+(aq) || Fe3+(aq) | Fe2+(aq) | Pt(s).A) –1.2 x 102 kJ/molB) –1.7 x 102 kJ/molC) 1.7 x 102 kJ/molD) –8.7 x 101 kJ/molE) –3.2 x 105 kJ/mol
A voltaic cell consists of an Mn/Mn2+ half-cell and a Pb/Pb2+ half-cell. Calculate [Pb2+] when [Mn2+] is 2.5 M and Ecell = 0.23 V.
A voltaic cell consists of an Mn/Mn2+ half-cell and a Pb/Pb2+ half-cell. Calculate [Pb2+] when [Mn2+] is 2.1 M and Ecell = 0.35 V. _____M
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!