A voltaic cell contains the following species involved in oxidation and reduction reactions: Au (s), Fe...
A voltaic cell is based on the reduction of Ag^+(aq) to Ag(s) and the oxidation of Sn(s) to Sn^2+(aq). (a) Write half-reactions for the cell's anode and cathode. Include the phases of all species in the chemical equation. Anode Cathode (b) Write a balanced cell reaction. Include the phases of all species in the chemical equation.
What is the standard cell potential (Y) for the spontaneous voltaic cell formed from the given half-reactions? Reduction Half-Reaction Ered (V) Anode -0.41 Fe2+ (aq) + 2e Fe(s) Cl2 (8) +20 +2 (aq) Cathode 1.36
38. The following redox half reactions are combined in a voltaic cell. Which reaction occurs at the cathode and what is the Eceu? Fe2+(aq) + 2e → Fe(s) E°=-0.44 V Cu²+(aq) + 2e → Cu(s) E°= 0.34 V a) b) c) d) Cu2+(aq) + 2e → Cu(s), Ecell = 0.78 V Fe2+(aq) + 2e → Fe(s), Ecel = 0.78 V Fe2+(aq) + 2e → Fe(s), Ecell =-0.10 V Cu²+(aq) + 2e → Cu(s), Ecel = 0.10 V Cu²+ (aq) +...
A galvanic cell is based on the following half-reactions: Fe2+ + 2e-→ Fe(s) Eº = -0.440 V 2H+ + 2e-→ H2(g) Eº = 0.000 V where the iron compartment contains an iron electrode and [Fe2+] = 1.00 × 10-3 M and the hydrogen compartment contains a platinum electrode, PH2 = 1.00 atm , and a weak acid, HA, at an initial concentration of 1.00 M. If the observed cell potential is 0.320 V at 25ºC, calculate the Ka value for...
Question 5 (Q3) The cell notation for a voltaic cell with the following redox reaction is: Fe2+ (aq) + 2e → Fe(s) (oxidation) » Cr3+ (aq) + 3e" (reduction) Cr(s) Cr3+ (aq) | Cr (s) "Fe (s) | Fe2+ (aq) Cr3+ (aq) | Cr (s) " Fe2+ (aq) | Fe (s) Fe2+ (aq)| Fe (s) "Cr (s) [Cr3+ (aq) II Cr(s) | Cr3+ (aq) Fe2+ (aq)| Fe (s) None of the above
1. A voltaic cell is constructed from a Al/Al half-cell (E'red -1.66 V) and a Fe/Fe? half- cell (Ered-0.44V) Give the oxidation half reaction and E"ox, the reduction half reaction and E'red, and the overall reaction and E"cell: Oxidation: Reduction: red E cell Overall Reaction: 2. For the cell Fe(s) Fe2+ | Pb* | Pb), the standard cell potential is +0.31 V. A cell using these reagents was made, and the observed potential was +0.37 V at 25°C. A) What...
Calculate the value of K given the following information anodel:(oxidation) Fe(s) Fe2+ (aq) + 2e E ? Fe = -0.447 V cathodel:(reduction): 2 x (Ag+ (aq) + e- Ag(s) Eig'lag = 0.7996; V 0.0592 Hint: Use E cell logk Calculate E Cell first. n 2.6 O 1.247 O 1.3X1042 none of the above
Design a voltaic cell with the following two reduction half -reactions : Design a voltaic cell with the following two reduction half-reactions Ag+(aq) + e-→ Ag(s) Zn2+(aq) + 2 e-→Zn(s) E 0.80 V Eo = 0.76 V Calculate Eocell and the equilibrium constant K for th copy of Final Exam cover sheet. e voltaic cell at 298 K. Click here for a E° cell -0.04 V and K-4.7 E cell-0.04 V and K-0.21 O Eocel =-0.04 V and K =...
PART II. Voltaic Cell Design In each of the following two problems, there's a pair of half-reactions shown in standard reduction form. In each problem, reverse the appropriate half-reaction, and indicate the following at the diagrams: a) voltage displayed by voltmeter (assume standard conditions) b) content of solution in each half-cell c) substance used for each electrode d) which electrode is anode, which is cathode e) which electrode is (+), which is (-) f) direction of electron flow g) direction...
A voltaic cell contains two half-cells. One half-cell contains a gold electrode immersed in a 1.00 M Au(NO3)3 solution. The second half-cell contains a magnesium electrode immersed in a 1.00 M Mg(NO3)2 solution. Au ** (aq) + 3 e Au(s) Ered = +1.498 V Mg2+ (aq) + 2 + Mg(s) Ered = -2.372 V (a) Using the standard reduction potentials given above, predict the standard cell potential of the voltaic cell. (b) Write the overall balanced equation for the voltaic...