For the cell schematic below, identify values for n and Q, and calculate the cell potential, Ecell.
Al(s)│Al3+(aq,0.15M)║Cu2+(aq,0.025M)│Cu(s)
At anode :
Oxidation : Al (s) --------------> Al3+ (aq) + 3e-
At cathode :
Reduction : Cu2+ (aq) + 2e- -------------> Cu (s)
Overall reaction :
2 Al (s) + 3 Cu2+ (aq) ------------> 2 Al3+ (aq) + 3 Cu (s)
Eocell = Eored - Eooxidation
= 0.34 - (- 1.66)
Eocell = 2.00 V
Q = [Al3+]^2 / [Cu2+]^3
= (0.15)^2 / (0.025)^3
Q = 1440
Q = 1.44 x 10^3
number of electrons , n = 6
Ecell = Eo - 0.05196 / 6 log Q
= 2.00 - 0.05196 / 6 log (1440)
Ecell = 1.97 V
For the cell schematic below, identify values for n and Q, and calculate the cell potential,...
22) Given the following redox reaction conducted in an acidic electrochemical cell, calculate ΔG®, ΔG, Eocell, and Ecell using the conditions provided. Al (s)VO2 (aq)Al (aqVo2 (aq) VO2(a)2H (aq)e VO2(a)H20() E 1.00V Al3+(aq) + 3e - Al (s) Eo 1.66V [VO2]- 1.2M [Al+0.025M vo21 0.05M [H ] 2.1M 22) Given the following redox reaction conducted in an acidic electrochemical cell, calculate ΔG®, ΔG, Eocell, and Ecell using the conditions provided. Al (s)VO2 (aq)Al (aqVo2 (aq) VO2(a)2H (aq)e VO2(a)H20() E 1.00V...
7) Calculate the cell potential of the following cell at 25°C. (7 points) Fe(s) Fe2+(aq) (1.1 M) || Cu2+ (aq) (0.50 M) Cu(s) Ecell = Eºcell - 0.0592/n logQ
Given the following redox reaction conducted in an acidic electrochemical cell, calculate ΔGo, ΔG, Eocell, and Ecell using the conditions provided. Al (s) + VO2+ (aq) → Al3+ (aq) + VO2+ (aq) VO2+(aq) + 2H+(aq) + e- → VO2+(aq) + H2O(l) Eo = 1.00V Al3+(aq) + 3e- → Al (s) Eo = -1.66V [VO2+] = 1.2M [Al3+] = 0.025M [VO2+] = 0.05M [H+] = 2.1M
Consider the following electrochemical cell: Al (s) I Al3+ (aq) (1.00 M) II Cu2+ (aq) (0.0020 M) I Cu (s) where Cu2+ aq + 2e- -> Cu (s) +0.34 V and Al3+ aq + 3e- -> Al (s) -1.66 V Calculate the standard cell potential for the given cell, calculate the cell potential for the given cell, and sketch the electrochemical cell using two beakers and labeling the electrodes, the cathode, the anode, the direction of electron flow in the...
8. The standard cell potential (E°cell) for the reaction below is +1.10 V. Calculate the cell potential for this reaction when the concentration of [Cu2+] = 1.0 × 10-5 M and [Zn2+] = 3.5 M. Zn (s) + Cu2+ (aq) → Cu (s) + Zn2+ (aq)
Conceptual: Consider a Sn(s)|Sn2+(aq) || Cu2+(aq)|Cu(s) cell. If the Sn2+ concentration is increased, what will happen to the measured Ecell value? • Calculation, full-reaction Nernst equation: Use the full Nernst equation to calculate Ecell for the conditions described… • Easier: Ni | Ni2+(0.300 M) || Cu2+(0.002 M) | Cu • Harder: Al | Al3+(0.002 M) || Cu2+(4.00 M) | Cu • Calculation, half-reaction Nernst equation: Use the Nernst equation to calculate E at pH 3.00 and [Cl- ] = 0.0035...
Calculate the cell potential for the reaction below when the concentrations of ions are: [Ag+ ] = 0.010 M and [Cu2+] = 0.750 M, at 25 °C. Cu(s) + 2Ag+ (aq) --> Cu2+(aq) + 2Ag(s) Given the standard reduction potentials: Cu2+(aq) + 2e– → Cu(s) Eϴ = 0.34 V Ag+ (aq) + e– → Ag(s) Eϴ = 0.80 V (A) 0.35 V (B) 0.44 V (C) 0.46 V (D) 0.48 V (E) 0.57 V
The cell potential of the following electrochemical cell depends on the pH of the solution in the anode half-cell: Pt(s)|H2(g,1atm)|H+(aq,?M)||Cu2+(aq,1.0M)|Cu(s). What is the pH of the solution if Ecell is 360 mV ?
Calculate the Ecell value at 298 K for the cell based on the reaction: Cu(s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag(s) where [Ag+] = 0.00350 M and [Cu2+] = 7.00x10-4 M. The standard reduction potentials are shown below: Ag+(aq) +e → Ag(s) E° = 0.7996 V Cu2+ (aq) + 2e -→ Cu(s) E° = 0.3419 V 2nd attempt Ecell = V
calculate the cell potential of this reaction. Cu|Cu2+ (0.0100M)||Cu2+ (0.1M)|Cu . The half-cell reaction is Cu2+(aq) + 2e- --> Cu(s) with standard potential Eo is 0.3419V