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A galvanic cell based on the following reactions Cu 2+ (aq) + 2e- + Cu(s) E°=0.339...
please answer all wueston Question 33 (3 points) (33) For a galvanic cell notation: (-) Ni...
please answer all wueston
Question 33 (3 points) (33) For a galvanic cell notation: (-) Ni /Ni2+ (aq) // Au3+ (aq) / Au (+), The electrode potentials: Eº (Ni2+/ Ni) = -0.257 V, E ° (AU3+ / Au ) = 1.498 V. The cell potential Eºcell (a) 1.241 V (b) - 1.755 V O (0) 1.755 V (d) 1.498 Why Question 35 (3 points) (35) For galvanic cell: (-) Cr/ Cr3+ (aq) / Cu2+ (aq) / Cu(+), the correct half...
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(aq,...
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(aq, 0.500 M) || Cu2+(aq, 1.00 M) Cu(s) Candidate 3: Mg(s) | Mg2+(aq, 0.500 M) | Pb2+(aq, 1.00 M)| Pb(s) (a) 6 pts) Choose one of the candidate voltaic cells #1, #2, or #3. Draw a schematic cell diagram for the candidate voltaic cell of choice. Clearly label anode, cathode, electrodes, ions and their concentrations, salt bridge, and the flow of electrons. (b) (5 pts)...
need answers for all questions Question 35 (3 points) (35) For galvanic cell: (-) Cr/ Cr3+...
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...
Give the balanced cell equation and determine e' for the galvanic cells based on the following...
Give the balanced cell equation and determine e' for the galvanic cells based on the following half-reactions. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.) H,02 + 2H+ 2e + 2H2O € = 1.78 V Cr2O72- + 14H+ +6e + 2 Cr3+ + 7H0 €* = 1.33 V a. V b. 2H+ + 2e + H2 € = 0.00 V A18+ +...
Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe (s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq...
Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe (s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq) E° (V) -0.74 -0.440 +0.771 +0.154 3Sn** (aq) +2Cr(s) → 2Cr** (aq) +3Sn²+ (aq) 3. What is the cell potential, Ecell, for the reaction above if [Sn] = 1.00 M, [Cr3+1 = 0.0200 M and [Sn2+] = 0.0100 M?
Give the balanced cell equation and determine e for the galvanic cells based on the following...
Give the balanced cell equation and determine e for the galvanic cells based on the following half-reactions. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.) H2O2 + 2H+ + 2 + 2H2O € = 1.78 V Cr2O72- + 14H+ +6e + 2 Cr3+ + 7H,0 6° = 1.33 V a. V + + V+ b. 2H+ + 2e → H€ = 0.00...
Consider the 2 half-reactions: Cu2+(aq) + 2e- --> Cu(s) Eo= 0.34V Ag+(aq) + e- --> Ag(s)...
Consider the 2 half-reactions: Cu2+(aq) + 2e- --> Cu(s) Eo= 0.34V Ag+(aq) + e- --> Ag(s) Eo= 0.8V If you increase the concentration of Cu2+(aq), which of the following would be true about the cell potential E? It would remain constant It would increase It would decrease This cannot be determined
Given the following half reactions what is the voltage of the galvanic cell that would result...
Given the following half reactions what is the voltage of the galvanic cell that would result from their combination [maximum positive voltage and spontaneous reaction). Show work. E° (volts) -0.74 Cr3+ (aq) + 3e Cu2+ (aq) + 2e → → Cr (8) Cu (s) +0.34 And what is the overall reaction that results after the half reactions are combined to cancel all electrons. Show work. And fill in the following abbreviated cell to represent the actual galvanic cell from above.
Consider a galvanic cell based on the following half reactions: E° (V) Mg2+ + 2e +...
Consider a galvanic cell based on the following half reactions: E° (V) Mg2+ + 2e + Mg -2.37 Cr3+ + 3e → Cr -0.73 What number of electrons are transferred per unit of cell reaction? Submit Hide Hints Hint 1 You must balance the cell reaction to determine the number of electrons transferred. This is the number that appears in the two half reactions after you have multiplied them by the appropriate integers so that when they are added together...
Consider a galvanic cell where one compartment has Br2 (aq) at a concentration of 0.50 M and Br ̄ at a concentration of...
Consider a galvanic cell where one compartment has Br2 (aq) at a concentration of 0.50 M and Br ̄ at a concentration of 0.10 M with a platinum electrode and the second compartment has Cr3+ at a concentration of 0.20 M and a Cr electrode. Use the following reduction potentials to answer the questions: Cr3+ (aq) + e ̄→Cr(s) εo = - 0.73 V Br2(aq) + 2e ̄→2Br ̄ (aq) εo =1.09V a. What is the standard cell potential for...
please answer all wueston
Question 33 (3 points) (33) For a galvanic cell notation: (-) Ni /Ni2+ (aq) // Au3+ (aq) / Au (+), The electrode potentials: Eº (Ni2+/ Ni) = -0.257 V, E ° (AU3+ / Au ) = 1.498 V. The cell potential Eºcell (a) 1.241 V (b) - 1.755 V O (0) 1.755 V (d) 1.498 Why Question 35 (3 points) (35) For galvanic cell: (-) Cr/ Cr3+ (aq) / Cu2+ (aq) / Cu(+), the correct half...
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(aq, 0.500 M) || Cu2+(aq, 1.00 M) Cu(s) Candidate 3: Mg(s) | Mg2+(aq, 0.500 M) | Pb2+(aq, 1.00 M)| Pb(s) (a) 6 pts) Choose one of the candidate voltaic cells #1, #2, or #3. Draw a schematic cell diagram for the candidate voltaic cell of choice. Clearly label anode, cathode, electrodes, ions and their concentrations, salt bridge, and the flow of electrons. (b) (5 pts)...
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...
Give the balanced cell equation and determine e' for the galvanic cells based on the following half-reactions. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.) H,02 + 2H+ 2e + 2H2O € = 1.78 V Cr2O72- + 14H+ +6e + 2 Cr3+ + 7H0 €* = 1.33 V a. V b. 2H+ + 2e + H2 € = 0.00 V A18+ +...
Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe (s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq) E° (V) -0.74 -0.440 +0.771 +0.154 3Sn** (aq) +2Cr(s) → 2Cr** (aq) +3Sn²+ (aq) 3. What is the cell potential, Ecell, for the reaction above if [Sn] = 1.00 M, [Cr3+1 = 0.0200 M and [Sn2+] = 0.0100 M?
Give the balanced cell equation and determine e for the galvanic cells based on the following half-reactions. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.) H2O2 + 2H+ + 2 + 2H2O € = 1.78 V Cr2O72- + 14H+ +6e + 2 Cr3+ + 7H,0 6° = 1.33 V a. V + + V+ b. 2H+ + 2e → H€ = 0.00...
Given the following half reactions what is the voltage of the galvanic cell that would result from their combination [maximum positive voltage and spontaneous reaction). Show work. E° (volts) -0.74 Cr3+ (aq) + 3e Cu2+ (aq) + 2e → → Cr (8) Cu (s) +0.34 And what is the overall reaction that results after the half reactions are combined to cancel all electrons. Show work. And fill in the following abbreviated cell to represent the actual galvanic cell from above.
Consider a galvanic cell based on the following half reactions: E° (V) Mg2+ + 2e + Mg -2.37 Cr3+ + 3e → Cr -0.73 What number of electrons are transferred per unit of cell reaction? Submit Hide Hints Hint 1 You must balance the cell reaction to determine the number of electrons transferred. This is the number that appears in the two half reactions after you have multiplied them by the appropriate integers so that when they are added together...
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