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I need help with questione 1-12 and discussion question 1 and 2. The previous pictures help...
1. Voltage reading for Fe-Cu cell 0.60 V 2. Voltage reading for Fe-Zn cell 0.25 V 3. Voltage reading for Cu-Zn cell 0.80 V 4. Fe2+ reduction half-reaction and Esedº Ered 5. Cu2+ reduction half-reaction and Ered Ered 6. Zn2+ reduction half-reaction and Ered Ered 7. Redox reaction of Fe-Cu cell 8. Fe-Cu Ecelº Ecell = 9. Redox reaction of Fe-Zn cell 10. Fe-Zn Ecetº Ece 11. Redox reaction of Cu-Zn cell 12. Cu-Zn Eceitº Ecelº = Questions to consider...
Data Sheet with Sample Data 1. Voltage reading for Fe-Cu cell 0.60 V 2. Voltage reading for Fe-Zn cell 0.25 V 3. Voltage reading for Cu-Zn cell 0.80 V 4. Fe2+ reduction half-reaction and Eredº Ered = 5. Cu2+ reduction half-reaction and Ered Eredº 6. Zn2+ reduction half-reaction and Ered Ered- 7. Redox reaction of Fe-Cu cell 8. Fe-Cu Ecelº Ecetº = 9. Redox reaction of Fe-Zn cell 10. Fe-Zn Ecello Ecelº = 11. Redox reaction of Cu-Zn cell 12....
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Please answer all parts that you can.
1. Which electrochemical cell had the greatest voltage? Identify the anode and the cathode for this pair, the measured cell potential, and the calculated Eºcell- 2. Which electrochemical cell had the smallest voltage? Identify the anode and the cathode for this pair, the measured cell potential, and the calculated Eºcell- 3. If the oxidation and reduction half-reactions are separated in a battery, this means the oxidizing agent is...
need help for half cell potentials pls calculate step by step
(NOTE - Remember that the positive electrode is attached to the red wire and the negative electrode is attached to the black wire.) Electrode Systems Used Anode (oxidation) Negative Cathode (reduction) Positive Measured Potential (V) Positive (Ecu) Copper and silver Cu() - Cu2+ + 2e" Ag+ +1e Ag) 0.432 V Zinc + Silver Zn cs + 2n**+ Zé dat + leº nAg (s) 1.484 V Copper & Zinc Zn(s)...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...
help did i do this correct?
a ws4 wws C PRE-L AX I Mycs aws4 Panapowe LUCY Name Electrochemical Cell Potentials and the Activity Series Report Form All data and calculations MUST be in your la notebook Part A. Electrochemical Cell Potentials Measured Potentials: NOTE - Remember that the positive electrode is attached to the red wire and the negative electrode is attached to the black wire.) Electrode Systems Used Anode (oxidation) Negative Cathode (reduction) Positive Measured Potential (V) an)...
Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g) Express your answer using two significant figures. Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V) Half-reaction E∘ (V) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) −0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) −0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) −0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) −0.76 Cu2+(aq)+2e−→Cu(s) 0.34 Mn2+(aq)+2e−→Mn(s) −1.18 2H+(aq)+2e−→H2(g) 0.00 Al3+(aq)+3e−→Al(s) −1.66 Fe3+(aq)+3e−→Fe(s) −0.036 Mg2+(aq)+2e−→Mg(s) −2.37 Pb2+(aq)+2e−→Pb(s) −0.13 Na+(aq)+e−→Na(s) −2.71 Sn2+(aq)+2e−→Sn(s) −0.14 Ca2+(aq)+2e−→Ca(s) −2.76 Ni2+(aq)+2e−→Ni(s) −0.23 Ba2+(aq)+2e−→Ba(s) −2.90 Co2+(aq)+2e−→Co(s) −0.28 K+(aq)+e−→K(s) −2.92 Cd2+(aq)+2e−→Cd(s)...
need help with the rest of the table
EXPERIMENT 10 DETERMINATION OF THE ELECTROCHEMICAL SERIES PURPOSE To determine the standard cell potential values of several electrochemical coll INTRODUCTION The basis for an electrochemical cell is an oxidation reduction Corredor be divided into two half reactions reaction. This reaction can Oxidation half reaction Gloss of electrons) takes place at the anode, which is the positive electrode that the anions migrate to Chence the name anode) Reduction half reaction (gain of electrons)...
Using the information in the table:
Which combination of metals, if used to create an
electrochemical cell, would produce the largest voltage?
Liu lur the reaction between Zn and Cu2+ ions is 1.1030 V, we can use the known value for the half-cell potential for zinc to determine the half-cell potential for copper: Zn(s) → Zn2+(aq) + 2e + Cu2+(aq) + 2e → Cu(s) Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) E half-cell = 0.7628 V Eºhalf-cell =...
Cell Potential and Equilibrium Standard reduction potentials The equilibrium constant, K, for a redox reaction is related to the standard cell potential, Ecel, by the equation Reduction half-reaction (V) Ag+ (aq) + e-→Ag(s) Cu2+ (aq) + 2e-→Cu(s) 0.34 Sn (a) 4e-Sn(s 0.15 2H' (aq) + 2e-→H2 (g) Ni2+ (aq) + 2e-→Ni(s)-0.26 Fe2+ (aq) + 2e-→Fe(s)-0.45 Zn2+ (aq) + 2e-→Zn(s)-0.76 Al3+ (aq) +3e-→Al(s) -1.66 Mg2+ (aq) + 2e-→Mg(s) -2.37 0.80 n FEcell where n is the number of moles of electrons...