Aluminum metal reacts with zinc(II) ion in an aqueous solution
by the following half-cell reactions:
Al(s) → Al3+(aq) +
3e−
Zn2+(aq) + 2e− → Zn(s)
a. Predict the potential of the cell under standard conditions. (4 points)
b. Predict whether the reaction will occur spontaneously, or whether a source of electricity will be required for the reaction. Justify your answer. (2 points)
c. In terms of the metals involved, predict the direction electrons will flow in the reaction. (2 points)
d. Predict which electrode will lose mass and which will gain mass. (2 points)
3. An electrochemical reaction occurs between an unknown element
and zinc. The half-cell reaction for the zinc is:
Zn(s) → Zn2+(aq) +
2e−
The cell potential for the reaction is Eºcell
= 1.83 V.
a. Is the reaction spontaneous or nonspontaneous? Explain. (1 point)
b. What is the half-cell potential of the unknown species? (2 points)
c. What is the identity of the unknown element? (2 points)
Aluminum metal reacts with zinc(II) ion in an aqueous solution by the following half-cell reactions: Al(s)...
Name: Chem 1120 Electrochemical Potentials Perform each of the following calculations, showing all work. Use the Electrochemical Potentials table provided on D2L if values are not provided. Standard State Electrochemical Cells 1. In each of the following systems two half-reactions are provided. For each system, a) write the balanced reaction occurring for a spontaneous system, and b) calculate the overall standard cell potential. a. Half Reaction Zn2+(aq) + 2e = Zn(s) Cr3+ (aq) + 38 = Cr(s) Eºred (V) -0.76...
In a copper-zinc voltaic cell, one half-cell consists of a ZnZn
electrode inserted in a solution of zinc sulfate and the other
half-cell consists of a CuCu electrode inserted in a copper sulfate
solution. These two half-cells are separated by a salt bridge.
At the zinc electrode (anode), ZnZn metal undergoes oxidation by
losing two electrons and enters the solution as Zn2+Zn2+ ions. The
oxidation half-cell reaction that takes place at the anode is
Zn(s)→Zn2+(aq)+2e−Zn(s)→Zn2+(aq)+2e−
The CuCu ions undergo reduction...
Half-cell Potentials: Half Reaction: E value +0.80 V Agt + e → Ag Fe3+ + € → Fe2+ +0.77 v +0.34 V -0.13 V Cu2+ +2e → Cu Pb2+ + 2e - → Ib Ni2+ + 2e → Ni Cd2+ +2e → Cd -0.25 V -0.40 V Fe2+ + 2e → Fe -0.44 V Zn2+ + 2e → Zn -0.76 V Al3+ +3e → AI - 1.66 V Consider an electrochemical cell constructed from the following half cells, linked by...
19 20 Question 16 Half-cell Potentials: Half Reaction: E' value + 0.80 V +0.77 V Agte → AS Fe3+ + + Fe2+ Cu2+ 2e → Cu Pb2+ + 2e → Pb +0.34 V -0.13 V Ni2+ + 2e → NI -0.25 V - 0.40 V Cd2+ +2e → ca Fe2+ + 2e → Fe Zn2+ + 2e → Zn -0.44 V - 0.76 V A13+ +3 → AI - 1.66 V Consider an electrochemical cell constructed from the following half...
A voltaic cell contains two half-cells. One half-cell contains a zinc electrode immersed in a 1.00 M Zn(NO3)2 solution. The second half-cell contains a titanium electrode immersed in a 1.00 M Ti(NO3)3 solution. Zn2+(aq) + 2 e− → Zn(s) E⁰red = −0.762 V Ti3+(aq) + 3 e− → Ti(s) E⁰red = −1.370 V (a) Using the standard reduction potentials given above, predict the standard cell potential of the voltaic cell. _____ V (b) Write the overall balanced equation for the...
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...
Draw the voltaic cell that will give the most positive cell potential choosing from the following half reactions: Eo (vs. SHE) Zn2+ (aq) + 2e- Zn (s), Eo= -0.76 Al3+ (aq) + 3e- Al (s), Eo= -1.66 Cr3+ (aq) + 3e- Cr (s), Eo= -0.74 Co2+ (aq) + 2e- Co (s), Eo= -0.28
need answers for all questions
Question 35 (3 points) (35) For galvanic cell: (-) Cr/ Cr3+ (aq) // Cu2+ (aq) / Cu (+), the correct half reactions are 0 (a) Cr3+ (a) Cr3+ + 3e → Crat (-) electrode; Cu+ 2e → Cu2+ at (+) electrode (b) Cr - 3e → Cr3+ at (-) electrode; Cu2+ + 2e → 2 Cu at (+) electrode Occ) cr (c) Cr + 3e → Cr3+ at (-) electrode; Cu2+ - 2e → Cu...
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)...
2. Calculate the standard cell potential for each of the following redox reactions, and then predict whether each will occur spontaneously as written. a. Sr(s) + Fe2+(aq) → Sr2+ (aq) + Fe(s) b. 2Cr(s) + 3Cd2+ (aq) → 2Cr3+ (aq) + 3Cd(s) 3. Calculate the standard cell potential, Eºcell, for each of the voltaic cells in Part II of the experiment. a. Zn(s) | Zn2+ (aq, 1.0 M) || Cu2+ (aq, 1.0 M) Cu(s) b. Zn(s) | Zn2+(aq, 1.0 M)...