Calculate the E of the cell based on the following half reactions: VO2+ (aq) + 2H+...
Calculate the E of the cell based on the following half reactions: VO2+ (aq) + 2H+ (aq) + e- ----> VO2+ (aq) + H2O (l)
Zn (s) --->Zn2+ (aq) + 2e-
Under the following condition: T = 25°C [VO2+] = 2.0 M [H+] = 0.5 M [VO2+] = 1.0 x 10-2 M [Zn2+] = 0.10 M
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Calculate the E of the cell based on the following half
reactions: VO2+ (aq) + 2H+...
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 ...
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 Part A Part complete An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.017M)+2H+(aq,1.3M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l) Calculate the cell potential under these nonstandard concentrations.
3. Calculate the cell potential for the voltaic cell based on the following half reactions at...
3. Calculate the cell potential for the voltaic cell based on the following half reactions at T = 25°C: Cr3+(aq) + 3e Cr(s) E° = -0.74 V TiO2(aq) + 2H(aq) + 1e → Ti+(aq) + H2O(1) E° = + 0.10 V Where, [Cr3+] = 1.0 x 104 M, [TiO2+] = 1.0 x 10-1M, [H+] = 1.0 M, [Ti$+] = 5.0 x 10-2 M.
A voltaic cell is based on the following half reactions at 25°C: Ag + e- →...
A voltaic cell is based on the following half reactions at 25°C: Ag + e- → Ag E = 0.8V H2O2 + 2H+ + 2e - → 2H20 E° = 1.78V Predict whether E cell is larger or smaller than Eºcel for the following cases. (a) (Ag +) = 1.0 M, (H2O2) = 2.0 M, [H*] = 2.0 M (b) (Ag +) - 2.0 M, (H2O2) = 1.0 M, [H) - 1.0 x 10'M
A voltaic cell is based on the following half reactions at 25 ⁰C: Ag + + e- à Ag &nbs...
A voltaic cell is based on the following half reactions at 25 ⁰C: Ag + + e- à Ag E⁰ = 0.8V H2O2 + 2H+ + 2e- à 2H2O E⁰ = 1.78V Predict whether E cell is larger or smaller than E⁰ cell for the following cases. Explain your answers. (a) [Ag +] = 1.0 M, [H2O2] = 2.0 M, [H+] = 2.0 M (b) [Ag +] = 2.0 M, [H2O2] = 1.0 M, [H+] = 1.0 x 10-7 M
Consider the galvanic cell based on the following half-reactions: Zn2+ + 2e + Zn * =...
Consider the galvanic cell based on the following half-reactions: Zn2+ + 2e + Zn * = -0.76 V Cd²+ + 2e - → Cd &* = -0.40 V a. Determine the overall cell reaction and calculate call (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.) V cell b. Calculate AG and K for the cell reaction at 25°C. AGE kJ K =...
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...
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...
In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of...
In a test of a new reference electrode, a chemist constructs a
voltaic cell consisting of a Zn/Zn2+ half-cell and an
H2/H+ half-cell under the following
conditions: [Zn2+ ] = 0.021 M [H+ ]= 1.3 M
partial pressure of H2 = 0.32 atm. Calculate
Ecell at 298 K (enter to 3 decimal places).
Zn2+ (aq) + 2e −
⟶ Zn(s) E° = − 0.76 V
2H+ (aq) + 2e −
⟶ H2(g) E° = 0.00 V
We were unable...
Consider a galvanic cell based on the following half reactions: E° (V) Zn2+ + 2e →...
Consider a galvanic cell based on the following half reactions: E° (V) Zn2+ + 2e → Zn -0.76 Au?+ + 3e → Au 1.50 If this cell is set up at 25°C with [Zn2] = 1.00 x 10-4M and [Au?') -2.00 10-2M, the expected cell potential is Submit Hide Hints Hint 1 Hint 2 The Nernst equation is based on the balanced cell reaction. Q reflects the mass action expression for the balanced cell reaction and n is the number...
Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2:...
Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2: Fe2+(aq)/Fe(s) and Zn2+(aq)/Zn(s) Which of the following is correct for the working cells? Standard reduction potentials, 298 K, Aqueous Solution (pH = 0): Cl2(g) + 2e --> 2C1-(aq); E° = +1.36 V Fe3+(aq) + e --> Fe2+(aq); E° = +0.77 V Cu2+(aq) + 2e --> Cu(s); E° = +0.34 V 2H+(aq) + 2e --> H2(g); E° = 0.00 V Pb2+(aq) + 2e --> Pb(s);...
In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of...
In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2 half- cell and an H2/H half-cell under the following conditions: [Zn2] = 0.042 M [H]- 19 M partial pressure of H2 =0.37 atm. Calculate Ecell at 298 K (enter to 3 decimal places). Zn2 (aq) + 2e +2H (aq) + 2e1 Eo-0.76 V E 0.00 V Zn(s) H2(g)
3. Calculate the cell potential for the voltaic cell based on the following half reactions at T = 25°C: Cr3+(aq) + 3e Cr(s) E° = -0.74 V TiO2(aq) + 2H(aq) + 1e → Ti+(aq) + H2O(1) E° = + 0.10 V Where, [Cr3+] = 1.0 x 104 M, [TiO2+] = 1.0 x 10-1M, [H+] = 1.0 M, [Ti$+] = 5.0 x 10-2 M.
A voltaic cell is based on the following half reactions at 25°C: Ag + e- → Ag E = 0.8V H2O2 + 2H+ + 2e - → 2H20 E° = 1.78V Predict whether E cell is larger or smaller than Eºcel for the following cases. (a) (Ag +) = 1.0 M, (H2O2) = 2.0 M, [H*] = 2.0 M (b) (Ag +) - 2.0 M, (H2O2) = 1.0 M, [H) - 1.0 x 10'M
Consider the galvanic cell based on the following half-reactions: Zn2+ + 2e + Zn * = -0.76 V Cd²+ + 2e - → Cd &* = -0.40 V a. Determine the overall cell reaction and calculate call (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.) V cell b. Calculate AG and K for the cell reaction at 25°C. AGE kJ K =...
In a test of a new reference electrode, a chemist constructs a
voltaic cell consisting of a Zn/Zn2+ half-cell and an
H2/H+ half-cell under the following
conditions: [Zn2+ ] = 0.021 M [H+ ]= 1.3 M
partial pressure of H2 = 0.32 atm. Calculate
Ecell at 298 K (enter to 3 decimal places).
Zn2+ (aq) + 2e −
⟶ Zn(s) E° = − 0.76 V
2H+ (aq) + 2e −
⟶ H2(g) E° = 0.00 V
We were unable...
Consider a galvanic cell based on the following half reactions: E° (V) Zn2+ + 2e → Zn -0.76 Au?+ + 3e → Au 1.50 If this cell is set up at 25°C with [Zn2] = 1.00 x 10-4M and [Au?') -2.00 10-2M, the expected cell potential is Submit Hide Hints Hint 1 Hint 2 The Nernst equation is based on the balanced cell reaction. Q reflects the mass action expression for the balanced cell reaction and n is the number...
Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2: Fe2+(aq)/Fe(s) and Zn2+(aq)/Zn(s) Which of the following is correct for the working cells? Standard reduction potentials, 298 K, Aqueous Solution (pH = 0): Cl2(g) + 2e --> 2C1-(aq); E° = +1.36 V Fe3+(aq) + e --> Fe2+(aq); E° = +0.77 V Cu2+(aq) + 2e --> Cu(s); E° = +0.34 V 2H+(aq) + 2e --> H2(g); E° = 0.00 V Pb2+(aq) + 2e --> Pb(s);...
In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2 half- cell and an H2/H half-cell under the following conditions: [Zn2] = 0.042 M [H]- 19 M partial pressure of H2 =0.37 atm. Calculate Ecell at 298 K (enter to 3 decimal places). Zn2 (aq) + 2e +2H (aq) + 2e1 Eo-0.76 V E 0.00 V Zn(s) H2(g)
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