Calculate Ecell (the reduction potential for the balanced chemical equation under nonstandard conditions) for the electrochemical cell shown below: Pt | Fe2+ (4.25x10-3 M), Fe3+ (1.50x10-3 M) || MnO4 - (6.50x10-3 M), Mn2+ (2.00x10-2 M), H+ (0.100 M) | Pt. Use the following standard reduction potentials for each half reaction. The Eo for Fe3+/Fe2+ = 0.77 V and Eo for MnO4 - /Mn2+ = 1.51 V.
Calculate Ecell (the reduction potential for the balanced chemical equation under nonstandard conditions) for the electrochemical...
A voltaic cell consists of an Al/Al3+ half-cell and an Fe2+/Fe3+ half-cell. Calculate Ecell when conc. of Fe2+= 0.402 M, Fe3+= 0.073 M and Al3+= 0.239 M. Use the reduction potentials for Al3+ is -1.66 V and for Fe3+ is 0.77 V. How do you decide which is oxi, red.
Show all steps please :) A galvanic cell is prepared using the following two half-cells: (i) MnO4 (0.273 M), Mn2+ (0.167 M), and H+ (1.0 M), and (ii) Fe2+ (0.247 M) and Fe3+ (0.389 M). The standard reduction potentials for the two half cells are: Mnoa + 8H+ + 5e – Mn2+ + 4H20 E° = 1.507 V Fe3+ + e- Fe2+ E° = 0.770 V a) Calculate the voltage for this galvanic cell. b) Write the balanced equation for...
Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C Cu(s) Cu2+ (0.15 M) Fe2+ (0.0039 M) Fe(s) E =-0.440 V E+Cu = 0.339 V Fe2+/Fe Is the electrochemical cell spontaneous or not spontaneous -0.779 Ecell = as written at 25 C? not spontaneous spontaneous о Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 'C Pt(s) Sn2 (0.0024 M), Sn4+ (0.12 M) |...
help me solve this correctly Calculate the value of E for the galvanic cell made from these two half reactions when MnO4) = 0.010 M. [H] =0.20 M, [Mn?] = 0.020 M, [Fe3+] = 0.75 M, [Fe2+] = 0.30 M. Ece - OTO nF 75 Q: 1.02m] 6.30m 5 (-010[0.1073747 (0.05 .000 4 MnO4 +8H+ +5e → Mn2+ + 4H20 E° = 1.51 V Fe3+ + + Fe2+ Ecelli Ecell - (8. (8./4.J/mol) (E° = 0.77 V Product S496485) Treactants...
Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C. Cu(s) Cu2+(0.14 M) | Fe2+(0.0044 M) Fe(s) Ecu?+Icu = 0.339 V Efez lfe = -0.440 V Ecell = v Is the electrochemical cell spontaneous or not spontaneous as written at 25 °C? O not spontaneous O spontaneous Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C. Pt(s) Sn2+(0.0048 M), Snº+(0.11 M) || Fe3+(0.12...
Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C Cu(s) Cu2 (0.12 M |Fe2 (0.0012 M) Fe(s) E2 =-0.440 V Efe/Fe = 0.339 V Cu2t/Cu Is the electrochemical cell spontaneous or not spontaneous Ecell V as written at 25 °C? not spontaneous spontaneous Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C. Pt(s) Sn2(0.0060 M), Sn4+(0.14 M) Fe3+(0.13 M), Fe2+(0.0056 M) Pt(s)...
Consider the following electrochemical cell: Pt | Cu2+ (aq) | Cu+ (aq) || Fe2+ (aq) | Fe3+ (aq)| Pt The cell is constructed by preparing one half-cell with a solution containing 8.33 * 10-3 M FeCl3 and 1.67*10-2 M FeCl2, and preparing the other half-cell with a solution containing 8.33*10-3 M CuCl2 and 0.025 M CuCl. Calculate the cell potential once the half-cells are connected to each other. E°(Cu2+/Cu+) = 0.16 V; E°(Fe2+/Fe3+) = 0.77 V
Electrochemistry - Equilibrium 1. The electrochemical cell described by the balanced chemical equation has a standard emf (electromotive force) of -0.08 V. Calculate the value (J) for the Wmax that the cell can do under standard conditions. Round your answer to 3 significant figures. St. Red. Pot. (V) Faraday's Constant Hg2+/Hg +0.85 F = 96485 C Fe3+/Fe2+ +0.77 2Fe3+(aq) + Hg(l) → 2Fe2+(aq) + Hg2+(aq) 2. The voltaic cell described by the balanced chemical equation has a standard emf of...
CHM 116 Homework #16 Due August 1, 2017 1. Calculate ΔGand K for the following electrochemical cell (Sample Exercise 17.7) (5 pts): Pt|Sn2+ (1M),Sn4+ (1M)||NO3 (1M),H+ (1M),NO(1atm)|Pt 2. Calculate ΔGand Eo for the following electrochemical cell at 98.0oC (Sample Exercise 17.7) (5 pts): Cu|Cu2+ (1M),||Zn2+ (1M)|Zn 3. Calculate E cell for the following electrochemical cell (Sample Exercise 17.6) (3 pts): Pt | Fe 2+ (0.0725 M), Fe3+ (0.110 M) | | MnO4 (0.0822 M), H+ (1.00 M), Mn2+ (0.0205 M)...
For the cell shown, the measured cell potential, Ecell, is -0.3711 V at 25 °C Pt(s) H2(g, 0.865 atm) | Ht (aq,? M) || Cd2+ (aq, 1.00 M) | Cd(s) The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, E°, are 2 H"(aq) + 2е — H2(g) Eo = 0.00 V Cd2+(aq)2 e - Cd(s) Eo =-0.403 V Calculate the Ht concentration М