Exam 19: Electrochemistry

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The Ecell for the following concentration cell at 25 °C is 0.006 V. Pt ∣ H2 (g,0.025 atm)∣ H+ (0.012 M) The E<sub>cell</sub> for the following concentration cell at 25 °C is 0.006 V. Pt ∣ H<sub>2</sub> (g,0.025 atm)∣ H<sup>+</sup> (0.012 M)    H<sup>+</sup> (1 M)∣ H<sub>2</sub> (g,1 atm)∣ Pt 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → H<sub>2</sub>(g)E° = 0.V H+ (1 M)∣ H2 (g,1 atm)∣ Pt 2 H+(aq)+ 2 e- → H2(g)E° = 0.V

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The net reaction in a voltaic cell with E°cell = +0.726 V is, 2 Fe3+(aq)+ 3 Zn(s)→ 2 Fe(s)+ 3 Zn2+(aq) What is ΔG° for this reaction at 25 °C?

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Consider the following reaction: 2 Al(s)+ 3 Ni2+(aq)→ 2 Al3+(aq)+ 3 Ni(s)E° = 1.410 V What would be the cell potential at 25 °C as calculated from the Nernst equation if [Ni2+] = 0.020 M,[Al3+] = 3.60 M?

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In the electrolysis of water,oxygen is evolved at the cathode.

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Choose the INCORRECT statement.

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The standard emf for the cell using the overall cell reaction below is +2.20 V: 2 Al(s)+ The standard emf for the cell using the overall cell reaction below is +2.20 V: 2 Al(s)+   (s)→   (aq)+   (aq) The emf generated by the cell when [   ] = 5.0 ×   Mol L<sup>-1</sup> and [   ] = 0.10 mol L<sup>-1</sup> is  (s)→ The standard emf for the cell using the overall cell reaction below is +2.20 V: 2 Al(s)+   (s)→   (aq)+   (aq) The emf generated by the cell when [   ] = 5.0 ×   Mol L<sup>-1</sup> and [   ] = 0.10 mol L<sup>-1</sup> is  (aq)+ The standard emf for the cell using the overall cell reaction below is +2.20 V: 2 Al(s)+   (s)→   (aq)+   (aq) The emf generated by the cell when [   ] = 5.0 ×   Mol L<sup>-1</sup> and [   ] = 0.10 mol L<sup>-1</sup> is  (aq) The emf generated by the cell when [ The standard emf for the cell using the overall cell reaction below is +2.20 V: 2 Al(s)+   (s)→   (aq)+   (aq) The emf generated by the cell when [   ] = 5.0 ×   Mol L<sup>-1</sup> and [   ] = 0.10 mol L<sup>-1</sup> is  ] = 5.0 × The standard emf for the cell using the overall cell reaction below is +2.20 V: 2 Al(s)+   (s)→   (aq)+   (aq) The emf generated by the cell when [   ] = 5.0 ×   Mol L<sup>-1</sup> and [   ] = 0.10 mol L<sup>-1</sup> is  Mol L-1 and [ The standard emf for the cell using the overall cell reaction below is +2.20 V: 2 Al(s)+   (s)→   (aq)+   (aq) The emf generated by the cell when [   ] = 5.0 ×   Mol L<sup>-1</sup> and [   ] = 0.10 mol L<sup>-1</sup> is  ] = 0.10 mol L-1 is The standard emf for the cell using the overall cell reaction below is +2.20 V: 2 Al(s)+   (s)→   (aq)+   (aq) The emf generated by the cell when [   ] = 5.0 ×   Mol L<sup>-1</sup> and [   ] = 0.10 mol L<sup>-1</sup> is

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Determine E°cell for the reaction: Pb2+(aq)+ Zn(s)→ Pb(s)+ Zn2+(aq).The half reactions are: Pb2+(aq)+ 2 e- → Pb(s)E° = -0.125 V Zn2+(aq)+ 2 e- → Zn(s)E° = -0.763 V

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Calculate Ecell at 25 °C for the following voltaic cell.Ksp for NiCO3(s)is 1.42 × 10-7. Ni(s)∣ Ni2+(aq)[sat'd NiCO3(s)] Calculate E<sub>cell </sub>at 25 °C for the following voltaic cell.K<sub>sp</sub> for NiCO<sub>3</sub>(s)is 1.42 × 10<sup>-7</sup>. Ni(s)∣ Ni<sup>2+</sup>(aq)[sat'd NiCO<sub>3</sub>(s)]    Ni<sup>2+</sup>(aq)(0.010M)∣ Ni(s) The half-reaction is: Ni<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Ni(s)E° = -0.257 V Ni2+(aq)(0.010M)∣ Ni(s) The half-reaction is: Ni2+(aq)+ 2 e- → Ni(s)E° = -0.257 V

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What is the shorthand notation that represents the following galvanic cell reaction? Pb(s)+ Cu(NO3)2(aq)→ Pb(NO3)2(aq)+ Cu(s)

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For the following voltaic cell,determine the [Cl-] when For the following voltaic cell,determine the [Cl<sup>-</sup>] when   = 0.500 atm,[Zn<sup>2+</sup>] = 1.77 × 10<sup>-2</sup> M,and E<sub>cell</sub> = 2.250 V.The half-reactions at 25 °C are: Cl<sub>2</sub>(g)+ 2 e<sup>-</sup> → 2 Cl<sup>-</sup>(aq)E° = +1.358 V Zn<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Zn(s)E° = -0.763 V Zn(s)∣ Zn<sup>2+</sup>(aq)   Cl<sup>-</sup>(aq),Cl<sub>2</sub>(g)∣ Pt(s) = 0.500 atm,[Zn2+] = 1.77 × 10-2 M,and Ecell = 2.250 V.The half-reactions at 25 °C are: Cl2(g)+ 2 e- → 2 Cl-(aq)E° = +1.358 V Zn2+(aq)+ 2 e- → Zn(s)E° = -0.763 V Zn(s)∣ Zn2+(aq) For the following voltaic cell,determine the [Cl<sup>-</sup>] when   = 0.500 atm,[Zn<sup>2+</sup>] = 1.77 × 10<sup>-2</sup> M,and E<sub>cell</sub> = 2.250 V.The half-reactions at 25 °C are: Cl<sub>2</sub>(g)+ 2 e<sup>-</sup> → 2 Cl<sup>-</sup>(aq)E° = +1.358 V Zn<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Zn(s)E° = -0.763 V Zn(s)∣ Zn<sup>2+</sup>(aq)   Cl<sup>-</sup>(aq),Cl<sub>2</sub>(g)∣ Pt(s) Cl-(aq),Cl2(g)∣ Pt(s)

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A solution of Al(NO3)3(aq)is electrolyzed by passing 144 A of current for 102 min.What mass of aluminum plates out?

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The standard Gibbs energy change for the following voltaic cell is ΔG° = -89.3 kJ at 25 °C.What is E°cell? The standard Gibbs energy change for the following voltaic cell is ΔG° = -89.3 kJ at 25 °C.What is E°<sub>cell</sub>?        Ag(s) The standard Gibbs energy change for the following voltaic cell is ΔG° = -89.3 kJ at 25 °C.What is E°<sub>cell</sub>?        Ag(s) The standard Gibbs energy change for the following voltaic cell is ΔG° = -89.3 kJ at 25 °C.What is E°<sub>cell</sub>?        Ag(s) Ag(s)

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Determine E°cell for the reaction: 2 Al(aq)+ 3 Zn2+(aq)→ 2 Al3+(aq)+ 3 Zn(s).The half reactions are: Al3+(aq)+ 3 e- → Al(s)E° = -1.676 V Zn2+(aq)+ 2 e- → Zn(s)E° = -0.763 V

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Determine E°cell for the reaction: MnO4-(aq)+ 8 H+(aq)+ 5 Fe2+(aq)→ Mn2+(aq)+ 4 H2O(l)+ 5 Fe3+(aq).The half reactions are: Fe3+(aq)+ e- → Fe2+(aq)E° = +0.771 V MnO4-(aq)+ 8 H+(aq)+ 5 e- → Mn2+(aq)+ 4 H2O(l)E° = 1.507 V

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In a galvanic cell,oxidation occurs at the cathode.

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Find Ecell at 25 °C for the following voltaic cell: Cu(s)∣ Cu2+(aq)(0.10 M) Find E<sub>cell</sub> at 25 °C for the following voltaic cell: Cu(s)∣ Cu<sup>2+</sup>(aq)(0.10 M)   Cu<sup>2+</sup>(aq)(0.85 M)∣ Cu(s) The half-reaction is: Cu<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Cu(s)E° = +0.377 volts Cu2+(aq)(0.85 M)∣ Cu(s) The half-reaction is: Cu2+(aq)+ 2 e- → Cu(s)E° = +0.377 volts

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Determine E°cell for the reaction: 2 Al3+(aq)+ 3 Zn(s)→ 2 Al + 3 Zn2+(aq).The half reactions are: Al3+(aq)+ 3 e- → Al(s)E° = -1.676 V Zn2+(aq)+ 2 e- → Zn(s)E° = -0.763 V

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Corrosion of metals is an oxidation-reduction process.

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Find Ecell for the following voltaic cell at 25 °C: Ti(s)∣ Ti2+(aq)(0.550M) Find E<sub>cell</sub> for the following voltaic cell at 25 °C: Ti(s)∣ Ti<sup>2+</sup>(aq)(0.550M)   Sn<sup>2+</sup>(aq)(0.005 M)∣ Sn(s) The half-reactions are: Ti<sup>2+</sup>(aq)+ 2 e<sup>-</sup> ⇔ Ti(s)E° = -1.630 V Sn<sup>2+</sup>(aq)+ 2 e<sup>-</sup> ⇔ Sn(s)E° = -0.137 V Sn2+(aq)(0.005 M)∣ Sn(s) The half-reactions are: Ti2+(aq)+ 2 e- ⇔ Ti(s)E° = -1.630 V Sn2+(aq)+ 2 e- ⇔ Sn(s)E° = -0.137 V

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Based on the following information, F2(g)+ 2 e- → 2 F-(aq)E° = +2.87 V Li +(aq)+ e- → Li(s)E° = -3.04 V Which of the following chemical species is the strongest reducing agent?

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