Exam 21: Electrochemistry: Chemical Change and Electrical Work

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Which of the following statements about voltaic and electrolytic cells is correct?

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In the shorthand notation for cells,a double vertical line is used to separate the reduced and oxidized forms of a redox couple.

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Calculate E°cell for the reaction of nickel(II)ions with cadmium metal at 25°C.K = 1.17 ×\times 105 Ni2+(aq)+ Cd(s) →\to Cd2+(aq)+ Ni(s)

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Which one of the following pairs of substances could be used to construct a single redox electrode (i.e. ,they have an element in common,but in different oxidation states)?

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A cell can be prepared from copper and tin.What is the E°cell for the cell that forms from the following half-reactions? Cu2+(aq)+ 2e- A cell can be prepared from copper and tin.What is the E°<sub>cell</sub> for the cell that forms from the following half-reactions? Cu<sup>2+</sup>(aq)+ 2e<sup>-</sup>   Cu(s);E° = 0.34 V Sn<sup>4+</sup>(aq)+ 2e<sup>-</sup>   Sn<sup>2+</sup>(aq);E° = 0.13 V Cu(s);E° = 0.34 V Sn4+(aq)+ 2e- A cell can be prepared from copper and tin.What is the E°<sub>cell</sub> for the cell that forms from the following half-reactions? Cu<sup>2+</sup>(aq)+ 2e<sup>-</sup>   Cu(s);E° = 0.34 V Sn<sup>4+</sup>(aq)+ 2e<sup>-</sup>   Sn<sup>2+</sup>(aq);E° = 0.13 V Sn2+(aq);E° = 0.13 V

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Consider the reaction CuO(s)+ H2(g) →\to Cu(s)+ H2O(l) In this reaction,which substances are the oxidant and reductant,respectively?

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Two cells are connected in series,so that the same current flows through two electrodes where the following half-reactions occur Cu2+(aq)+ 2e- →\to Cu(s)and Ag+(aq)+ e- →\to Ag(s) For every 1.00 g of copper produced in the first process,how many grams of silver will be produced in the second one?

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Examine the following half-reactions and select the strongest oxidizing agent among the substances. [PtCl4]2-(aq)+ 2e- Examine the following half-reactions and select the strongest oxidizing agent among the substances. [PtCl<sub>4</sub>]<sup>2-</sup>(aq)+ 2e<sup>-</sup>   Pt(s)+ 4Cl<sup>-</sup>(aq);E° = 0.755 V RuO<sub>4</sub>(s)+ 8H<sup>+</sup>(aq)+ 8e<sup>-</sup>   Ru(s)+ 4H<sub>2</sub>O(l);E° = 1.038 V FeO<sub>4</sub><sup>2-</sup>(aq)+ 8H<sup>+</sup>(aq)+ 3e<sup>-</sup>   Fe<sup>3+</sup>(aq)+ 4H<sub>2</sub>O(l);E° = 2.07 V H<sub>4</sub>XeO<sub>6</sub>(aq)+ 2H<sup>+</sup>(aq)+ 2e<sup>-</sup>   XeO<sub>3</sub>(aq)+ 3H<sub>2</sub>O(l);E° = 2.42 V Pt(s)+ 4Cl-(aq);E° = 0.755 V RuO4(s)+ 8H+(aq)+ 8e- Examine the following half-reactions and select the strongest oxidizing agent among the substances. [PtCl<sub>4</sub>]<sup>2-</sup>(aq)+ 2e<sup>-</sup>   Pt(s)+ 4Cl<sup>-</sup>(aq);E° = 0.755 V RuO<sub>4</sub>(s)+ 8H<sup>+</sup>(aq)+ 8e<sup>-</sup>   Ru(s)+ 4H<sub>2</sub>O(l);E° = 1.038 V FeO<sub>4</sub><sup>2-</sup>(aq)+ 8H<sup>+</sup>(aq)+ 3e<sup>-</sup>   Fe<sup>3+</sup>(aq)+ 4H<sub>2</sub>O(l);E° = 2.07 V H<sub>4</sub>XeO<sub>6</sub>(aq)+ 2H<sup>+</sup>(aq)+ 2e<sup>-</sup>   XeO<sub>3</sub>(aq)+ 3H<sub>2</sub>O(l);E° = 2.42 V Ru(s)+ 4H2O(l);E° = 1.038 V FeO42-(aq)+ 8H+(aq)+ 3e- Examine the following half-reactions and select the strongest oxidizing agent among the substances. [PtCl<sub>4</sub>]<sup>2-</sup>(aq)+ 2e<sup>-</sup>   Pt(s)+ 4Cl<sup>-</sup>(aq);E° = 0.755 V RuO<sub>4</sub>(s)+ 8H<sup>+</sup>(aq)+ 8e<sup>-</sup>   Ru(s)+ 4H<sub>2</sub>O(l);E° = 1.038 V FeO<sub>4</sub><sup>2-</sup>(aq)+ 8H<sup>+</sup>(aq)+ 3e<sup>-</sup>   Fe<sup>3+</sup>(aq)+ 4H<sub>2</sub>O(l);E° = 2.07 V H<sub>4</sub>XeO<sub>6</sub>(aq)+ 2H<sup>+</sup>(aq)+ 2e<sup>-</sup>   XeO<sub>3</sub>(aq)+ 3H<sub>2</sub>O(l);E° = 2.42 V Fe3+(aq)+ 4H2O(l);E° = 2.07 V H4XeO6(aq)+ 2H+(aq)+ 2e- Examine the following half-reactions and select the strongest oxidizing agent among the substances. [PtCl<sub>4</sub>]<sup>2-</sup>(aq)+ 2e<sup>-</sup>   Pt(s)+ 4Cl<sup>-</sup>(aq);E° = 0.755 V RuO<sub>4</sub>(s)+ 8H<sup>+</sup>(aq)+ 8e<sup>-</sup>   Ru(s)+ 4H<sub>2</sub>O(l);E° = 1.038 V FeO<sub>4</sub><sup>2-</sup>(aq)+ 8H<sup>+</sup>(aq)+ 3e<sup>-</sup>   Fe<sup>3+</sup>(aq)+ 4H<sub>2</sub>O(l);E° = 2.07 V H<sub>4</sub>XeO<sub>6</sub>(aq)+ 2H<sup>+</sup>(aq)+ 2e<sup>-</sup>   XeO<sub>3</sub>(aq)+ 3H<sub>2</sub>O(l);E° = 2.42 V XeO3(aq)+ 3H2O(l);E° = 2.42 V

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Cryolite,Na3AlF6 is used in the electrolysis of aluminum oxide because

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The voltaic cell made up of cobalt,copper and their M2+ ions,has E°cell = 0.62 V.If E° of the cathode half-cell is 0.34 V,what is E° of the anode half-cell? Cu2+(aq)+ Co(s) →\to Cu(s)+ Co2+(aq)

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The lead-acid battery is an example of a secondary battery.

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Consider the following balanced redox reaction 3CuO(s)+ 2NH3(aq) →\to N2(g)+ 3H2O(l)+ 3Cu(s) Which of the following statements is true?

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A voltaic cell consists of a Cd/Cd2+ electrode (E° = -0.40 V)and a Fe/Fe2+ electrode (E°= -0.44 V).If Ecell = 0 and the temperature is 25°C,what is the ratio [Fe2+]/[Cd2+]?

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A voltaic cell prepared using zinc and iodine has the following cell notation. Zn(s)| Zn2+(aq)|| I-(aq)| I2(s)| C (graphite) Which of the following equations correctly represents the balanced,spontaneous,cell reaction?

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Consider the non-aqueous cell reaction 2Na(l)+ FeCl2(s)  Consider the non-aqueous cell reaction 2Na(l)+ FeCl<sub>2</sub>(s)   2NaCl(s)+ Fe(s) For which E°<sub>cell</sub> = 2.35 V at 200°C. \Delta G° at this temperature is 2NaCl(s)+ Fe(s) For which E°cell = 2.35 V at 200°C. Δ\Delta G° at this temperature is

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Calculate Δ\Delta G° for the oxidation of 3 moles of copper by nitric acid. Cu2+(aq)+ 2e-  Calculate  \Delta G° for the oxidation of 3 moles of copper by nitric acid. Cu<sup>2+</sup>(aq)+ 2e<sup>-</sup>   Cu(s);E° = 0.34 V NO<sub>3</sub><sup>-</sup>(aq)+ 4H<sup>+</sup>(aq)+ 3e<sup>-</sup>   NO(g)+ 2H<sub>2</sub>O(l);E° = 0.957 V Cu(s);E° = 0.34 V NO3-(aq)+ 4H+(aq)+ 3e-  Calculate  \Delta G° for the oxidation of 3 moles of copper by nitric acid. Cu<sup>2+</sup>(aq)+ 2e<sup>-</sup>   Cu(s);E° = 0.34 V NO<sub>3</sub><sup>-</sup>(aq)+ 4H<sup>+</sup>(aq)+ 3e<sup>-</sup>   NO(g)+ 2H<sub>2</sub>O(l);E° = 0.957 V NO(g)+ 2H2O(l);E° = 0.957 V

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What mass of copper will be deposited when 18.2 A are passed through a CuSO4 solution for 45.0 minutes?

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A voltaic cell consists of an Au/Au3+ electrode (E° = 1.50 V)and a Cu/Cu2+ electrode (E° = 0.34 V).Calculate [Au3+] if [Cu2+] = 1.20 M and Ecell = 1.13 V at 25°C.

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Manganese dioxide (MnO2)for use in dry cells is made by electrolyzing solutions of Mn2+,where the following reaction occurs at the anode: Mn2+(aq)+ 2H2O(l) →\to MnO2(s)+ 4H+(aq)+ 2e- Using a current of 100.amperes (A),how many hours will it take to produce 5.00 kg of MnO2,according to the above reaction?

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Which component of the following cell notation is the anode? P | Q || R | S

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