Question 1

The electrochemical reaction that powers a lead-acid storage battery is as follows: Pb(s)+ PbO₂(s)+ 4 H⁺(aq)+ 2 SO₄^2-(aq)→ 2 PbSO₄(s)+ 2 H₂O( ) A single cell of this battery consists of a Pb electrode and a PbO₂ electrode,each submerged in sulfuric acid.Which of the following reactions occurs at the cathode during discharge? A) Pb(s)is reduced to PbSO₄(s). B) PbO₂(s)is reduced to PbSO₄(s). C) PbO₂(s)is oxidized to PbSO₄(s). D) Pb(s)is oxidized to PbSO₄(s). E) H⁺ is oxidized to H₂O( ).

Accepted Answer

During discharge, the PbO2 electrode (also called the positive electrode or the anode) undergoes reduction to form PbSO4 while the Pb electrode (also called the negative electrode or the cathode) undergoes oxidation to form PbSO4. Thus, the reaction at the cathode can be represented as: Pb(s) + HSO4- → PbSO4 + H+ + 2e-. Therefore, the correct choice is B because PbO2 is reduced at the anode and not at the cathode.

Question 2

Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb²⁺(aq)+ 2 e^- → Pb(s) E° = -0.126 V Fe³⁺(aq)+ e^- → Fe²⁺(s) E° = +0.771 V A) Pb²⁺(aq)+ 2 Fe²⁺(s)→ Pb(s)+ 2 Fe³⁺(aq); = +0.897 V B) Pb²⁺(aq)+ Fe²⁺(s)→ Pb(s)+ Fe³⁺(aq); = +0.645 V C) Pb(s)+ 2 Fe³⁺(aq)→ Pb²⁺(aq)+ 2 Fe²⁺(s); = +1.416 V D) Pb(s)+ 2 Fe³⁺(aq)→ Pb²⁺(aq)+ 2 Fe²⁺(s); = +0.897 V E) Pb(s)+ Fe³⁺(aq)→ Pb²⁺(aq)+ Fe²⁺(s); = +0.645 V

Accepted Answer

The overall reaction involves the reduction of PbS to Pb and the oxidation of FeS to FeS2. The half-reactions must be flipped and multiplied by factors to balance the number of electrons transferred.

PbS + 2e- -> Pb (E° = -0.126 V)
FeS -> FeS2 + 2e- (E° = -0.771 V)

Adding the two reactions gives:

PbS + FeS -> Pb + FeS2

The standard cell potential can be found using the formula:

E°cell = E°reduction + E°oxidation
E°cell = (-0.126 V) + (-0.771 V)
E°cell = -0.897 V

Choice D gives the correct overall reaction and standard cell potential.

Question 3

Write a balanced net ionic equation for the overall reaction represented by the following cell notation. Al(s)| Al³⁺(aq)|| Cl₂(g)| Cl-(aq)| Pt(s) A) 2 Al(s)+ 3 Cl₂(g)→ 2 Al³⁺(aq)+ 6 Cl^-(aq) B) Al(s)+ Al³⁺(aq)→ Cl^-(aq)+ Cl₂(g) C) 2 Al³⁺(aq)+ 6 Cl^-(aq)→ 2 Al(s)+ 3 Cl₂(g) D) Al(s)+ 3 Cl₂(g)→ Al³⁺(s)+ 2 Cl^-(aq) E) Al(s)+ 2 Cl^-(aq)→ Cl₂(g)+ Al³⁺(aq)

Accepted Answer

The correct net ionic equation involves aluminum solid reacting with chlorine gas to form aluminum ions in solution and chloride ions in solution, with the coefficients ensuring the charge and atoms are balanced.

Question 4

An SHE electrode has been assigned a standard reduction potential,E°,of 0.00 volts.Which of the following reactions will occur at this electrode? A) 2 H₂O( )+ 2 e^- → H₂(g)+ 2 OH^-(aq) B) O₂(g)+ 4 e^- → 2 O^2-(aq) C) Hg₂Cl₂(s)+ 2 e^- → 2 Hg( )+ 2 Cl^-(aq) D) Li⁺(aq)+ e^- → Li(s) E) 2 H⁺(aq)+ 2 e^- → H₂(g)

Accepted Answer

The answer of An SHE electrode has been assigned a...

Question 5

Calculate the standard cell potential ( )for the reaction 2 Ag(s)+ Co²⁺(aq)→ 2 Ag⁺(aq)+ Co(s). The standard reduction potentials are as follows: Ag⁺(aq)+ e⁻→ Ag(s) E° = 0.8 V Co²⁺(aq)+2 e⁻→ Co(s) E° = -0.277 V A) -1.077 V B) 1.077 V C) -1.877 V D) 1.877 V E) 1.323 V

Accepted Answer

The standard cell potential can be calculated using the formula:
E°cell = E°cathode - E°anode
where E°cathode is the standard reduction potential of the cathode (Ag), and E°anode is the standard reduction potential of the anode (Co).
Substituting the given values, we get:
E°cell = E°cathode - E°anode
E°cell = 0.8 V - (-0.277 V)
E°cell = 1.077 V
Therefore, the correct choice is A.

Question 6

In the given electrochemical cell,which of the following is the cathode half-reaction? Zn(s)| Zn²⁺(aq)|| Fe³⁺(aq),Fe²⁺(aq)| Pt(s) A) Fe³⁺(aq)+ e− → Fe²⁺(aq) B) Fe²⁺(aq)+ e− → Fe³⁺(aq) C) Fe²⁺(aq)+ Pt(s)→ Fe³⁺(aq)+ e− D) Zn²⁺(aq)→ Zn(s)+ 2 e− E) Zn(s)→ Zn²⁺(aq)+ 2 e−

Accepted Answer

The cathode is where reduction occurs, and option A correctly shows the reduction of Fe^3+ to Fe^2+, which is consistent with the behavior of a cathode in an electrochemical cell.

Question 7

Calculate E°_cell for the cell for the reaction 2 Ga(s)+ 3 Sn⁴⁺(aq)→ 3 Sn²⁺(aq)+2 Ga³⁺(aq). The standard reduction potentials are as follows: Ga³⁺(aq)+ 3 e− → Ga(s) E° = -0.55 V Sn⁴⁺(aq)+ 2 e− → Sn²⁺(aq) E° = 0.15V A) 0.70 V B) 0.40 V C) 0.33 V D) -0.40 V E) −0.70 V

Accepted Answer

The standard cell potential (E°cell) is calculated by subtracting the standard reduction potential of the anode (the more negative value) from the standard reduction potential of the cathode (the more positive value). Here, Ga is oxidized (anode reaction) and Sn2+ is reduced (cathode reaction). Thus, E°cell = 0.15 V (for Sn2+/Sn) - (-0.55 V) (for Ga3+/Ga) = 0.70 V.

Question 8

Use the following standard reduction potentials to determine which species is the strongest reducing agent. 2 H⁺(aq)+ 2 e^- → H₂(g); 0.00 V K⁺(aq)+ e^- → K(s); -2.93 V F₂(g)+ 2 e^- → 2 F^-(aq); 2.87 V Al³⁺(aq)+ 3 e^- → Al(s); -1.66 V Pb²⁺(aq)+ 2 e^- → Pb(s); -0.13 V A) Al³⁺ B) H⁺ C) Pb²⁺ D) F^- E) K

Accepted Answer

The strongest reducing agent is the one with the most negative standard reduction potential, which is potassium (K) with a potential of -2.93 V.

Question 9

Which of the following species are likely to behave as oxidizing agents? Li(s),H₂(g),MnO₄^-(aq),and Cl^-(aq) A) Li(s) B) MnO₄^-(aq) C) H₂(g)and Cl^-(aq) D) Li(s)and MnO₄^-(aq) E) Cl^-(aq)

Accepted Answer

MnO₄⁻(aq) is a strong oxidizing agent because it can accept electrons and be reduced to Mn²⁺.

Question 10

The unit for electromotive force,emf,is the volt.1 volt is equal to _____.
A) one joule per second
B) one coulomb per joule
C) one joule per coulomb
D) one coulomb per second
E) one calorie per second

Accepted Answer

The unit for emf, volt, is defined as the potential difference required to transfer one joule of energy per coulomb of charge. Therefore, one volt is equal to one joule per coulomb.

Question 11

According to the cell notation below,which of the following species is undergoing reduction? Ni | Ni²⁺(aq)|| Mn²⁺(aq)| MnO₂(s)| Pt(s) A) Mn²⁺(aq) B) Ni²⁺(aq) C) Ni(s) D) MnO₂(s) E) Pt(s)

Accepted Answer

The answer of According to the cell notation below,which of...

Question 12

Use the following standard reduction potentials to determine which species is the strongest oxidizing agent. Fe²⁺(aq)+ 2 e^- → Fe(s); E° = -0.41 V Pt²⁺(aq)+ 2 e^- → Pt(s); E° = 1.18 V Cr₂O₇^2-(aq)+ 14 H⁺(aq)+ 6 e^- → 2 Cr³⁺(aq)+ 7 H₂O( ); E° = 1.33 V A) Fe B) Pt C) Cr³⁺ D) Fe²⁺ E) Cr₂O₇^2-

Accepted Answer

The answer of Use the following standard reduction potentials to...

Question 13

Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt²⁺(aq)+ 2 e− → Pt(s); E° = 1.180 V Pb²⁺(aq)+ 2 e− → Pb(s); E° = -0.130 V A) Pt²⁺(aq)+ Pb(s)→ Pt(s)+ Pb²⁺(aq); = 1.310 V B) Pt(s)+ Pb²⁺(aq)→ Pt²⁺(aq)+ Pb(s); = −1.310 V C) Pt²⁺(aq)+ Pb²⁺(aq)→ Pt(s)+ Pb(s); = 1.050 V D) Pt²⁺(aq)+ Pb(s)→ Pt(s)+ Pb²⁺(aq); =0.655 V E) Pt(s)+ Pb²⁺(aq)→ Pt²⁺(aq)+ Pb(s); = −0.655 V

Accepted Answer

The answer of Which of the following is the balanced...

Question 14

Consider the following half-reactions. Ag⁺(aq)+ e^- → Ag(s) E° = +0.80 V Cu²⁺(aq)+ 2 e^- → Cu(s) E° = +0.34 V Pb²⁺(aq)+ 2 e^- → Pb(s) E° = -0.13 V Fe²⁺(aq)+ 2 e^- → Fe(s) E° = -0.44 V Al³⁺(aq)+ 3 e^- → Al(s) E° = -1.66 V Which of the following species will oxidize lead,Pb(s)? A) Ag⁺(aq)and Cu²⁺(aq) B) Ag(s)and Cu(s) C) Fe²⁺(aq)and Al³⁺(aq) D) Fe(s)and Al(s) E) Cu²⁺(aq)and Fe²⁺(aq)

Accepted Answer

Species with a higher reduction potential (E°) can oxidize species with a lower reduction potential. Ag^+ (E° = +0.80 V) and Cu^2+ (E° = +0.34 V) have higher reduction potentials than Pb^2+ (E° = -0.13 V), meaning they can accept electrons from Pb(s), oxidizing it.

Question 15

Use the following standard reduction potentials to determine which species is the best oxidizing agent. O₂(g)+ 4 H⁺(aq)+ 4 e^- → 2 H₂O( ) E° = +1.229 V Hg₂²⁺(aq)+ 2 e^- → 2 Hg( ) E° = +0.789 V I₂(s)+ 2 e^- → 2 I^-(aq) E° = +0.535 V A) I₂(s) B) O₂(g) C) I^-(aq) D) Hg₂²⁺(aq) E) H₂O( )

Accepted Answer

The species with the highest E° value is the best oxidizing agent. Therefore, the best choice is O₂(g) with an E° value of +1.229 V.

Question 16

Consider the following half-reactions. Cl₂(g)+ 2 e^- → 2 Cl^-(aq) E° = +1.36 V Ag⁺(aq)+ e^- → Ag(s) E° = +0.80 V Cu²⁺(aq)+ 2 e^- → Cu(s) E° = +0.34 V Sn²⁺(aq)+ 2 e^- → Sn(s) E° = -0.14 V Al³⁺(aq)+ 3 e^- → Al(s) E° = -1.66 V Which of the following species will reduce Cu²⁺(aq)ion? A) Ag(s)and Sn²⁺(aq) B) Cl^-(aq)and Ag(s) C) Cl₂(g)and Ag⁺(aq) D) Sn(s)and Al(s) E) Sn²⁺(aq)and Al³⁺(aq)

Accepted Answer

The answer of Consider the following half-reactions. Cl₂(g)+ 2 e^-...

Question 17

Which of the following is the cell notation for a voltaic cell based on the following reaction? Cu²⁺(aq)+ Pb(s)+ SO₄^2-(aq)→ Cu(s)+ PbSO₄(s) A) Pb(s)| PbSO₄(s)|| Cu²⁺(aq)|| Cu(s) B) Cu(s)| Cu²⁺(aq)|| SO₄^2-(aq)| PbSO₄(s)| Pb(s) C) Cu(s)| Cu²⁺(aq),SO₄^2-(aq)| PbSO₄(s)| Pb(s) D) Cu(s)| Cu²⁺(aq),SO₄^2-(aq)|| PbSO₄(s)|| Pb(s) E) Pb(s)| PbSO₄(s)| SO₄^2-(aq)|| Cu²⁺(aq)|| Cu(s)

Accepted Answer

In a voltaic cell, the anode (oxidation) is on the left and the cathode (reduction) is on the right. The correct notation is Pb(s)| PbSO₄(s)| SO₄²⁻(aq)|| Cu²⁺(aq)|| Cu(s), where Pb is oxidized and Cu²⁺ is reduced.

Question 18

Write a balanced net ionic equation for the overall reaction represented by the following cell notation. Cu(s)| Cu²⁺(aq)|| Mn²⁺(aq)| Mn(s) A) 2 Cu(s)+ Mn²⁺(aq)→ Mn(s)+ 2 Cu²⁺(aq) B) Cu(s)+ Cu²⁺(aq)→ Mn(s)+ Mn²⁺(aq) C) Cu(s)+ Mn²⁺(aq)→ Mn(s)+ Cu²⁺(aq) D) 2 Mn(s)+ Cu²⁺(aq)→ Cu(s)+ 2Mn²⁺(aq) E) Mn(s)+ Cu²⁺(aq)→ Cu(s)+ Mn²⁺(aq)

Accepted Answer

The cell notation shows the reactants and products in the half-cells. The left half-cell has copper metal and copper sulfate solution, while the right half-cell has manganese metal and manganese sulfate solution. The anode reaction (left half-cell) is oxidation of copper metal to Cu2+ ions, which react with sulfate ions to form CuSO4 solution. The cathode reaction (right half-cell) is reduction of Mn2+ ions to manganese metal. The net ionic equation for the entire reaction is Cu(s) + Mn2+(aq) → Cu2+(aq) + Mn(s). However, the given answer choices have different combinations of reactants and products. Among them, choice C represents the net ionic equation after cancelling out the spectator ions (SO42- and Na+ if present) and putting the remaining ions in the correct products and reactants. Therefore, choice C is the best choice.

Question 19

Which of the following is the cell notation for a cell in which the hydrogen electrode is the anode and the cathode half-reaction is Co³⁺(aq)+ e− → Co²⁺(aq)? A) Pt(s)| H₂(g)| H⁺(aq)|| Co³⁺(aq),Co²⁺(aq)| Pt(s) B) Pt(s)| H₂(g)| H⁺(aq)|| Co³⁺(aq),Co²⁺(aq) C) Co²⁺(aq),Co³⁺(aq)|| H⁺(aq)| H₂(g)| Pt(s) D) Pt(s)| Co²⁺(aq),Co³⁺(aq)|| H⁺(aq)| H₂(g)| Pt(s) E) H₂(g)| H⁺(aq)|| Co²⁺(aq),Co³⁺(aq)

Accepted Answer

The anode is the hydrogen electrode, which is represented as Pt(s)| H2(g)| H+(aq). The cathode half-reaction is Co³⁺(aq)+ e− → Co²⁺(aq), which can be rearranged as Co²⁺(aq)| Co³⁺(aq). Therefore, the cell notation is Pt(s)| H₂(g)| H⁺(aq)|| Co³⁺(aq),Co²⁺(aq)| Pt(s).

Question 20

Which of the following overall chemical equations is responsible for generating electricity in fuel cells used in NASA's Space Shuttle programs? A) Fe₂O₃(s)+ Al(s)→ Fe(s)+ Al₂O₃(s) B) Pb(s)+ PbO₂(s)+ 2 H₂SO₄(aq)→ 2 PbSO₄(s)+ 2 H₂O( ) C) 2 NiO(OH)(s)+ Cd(s)+ H₂O( )→ 2 Ni(OH)₂(s)+ Cd(OH)₂(s) D) 2 H₂(g)+ O₂(g)→ 2 H₂O( ) E) N₂H₄( )+ O₂(g)→ N₂(g)+ 2 H₂O( )

Accepted Answer

The answer of Which of the following overall chemical equations...

Quiz 20: Principles of Chemical Reactivity: Electron Transfer Reactions

Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb²⁺(aq) + 2 e^- → Pb(s) E° = -0.126 V Fe³⁺(aq) + e^- → Fe²⁺(s) E° = +0.771 V

Write a balanced net ionic equation for the overall reaction represented by the following cell notation. Al(s) | Al³⁺(aq) || Cl₂(g) | Cl-(aq) | Pt(s)

An SHE electrode has been assigned a standard reduction potential,E°,of 0.00 volts.Which of the following reactions will occur at this electrode?

Calculate the standard cell potential ( ) for the reaction 2 Ag(s) + Co²⁺(aq) → 2 Ag⁺(aq) + Co(s) . The standard reduction potentials are as follows: Ag⁺(aq) + e⁻→ Ag(s) E° = 0.8 V Co²⁺(aq) +2 e⁻→ Co(s) E° = -0.277 V

In the given electrochemical cell,which of the following is the cathode half-reaction? Zn(s) | Zn²⁺(aq) || Fe³⁺(aq) ,Fe²⁺(aq) | Pt(s)

Calculate E°_cell for the cell for the reaction 2 Ga(s) + 3 Sn⁴⁺(aq) → 3 Sn²⁺(aq) +2 Ga³⁺(aq) . The standard reduction potentials are as follows: Ga³⁺(aq) + 3 e− → Ga(s) E° = -0.55 V Sn⁴⁺(aq) + 2 e− → Sn²⁺(aq) E° = 0.15V

Use the following standard reduction potentials to determine which species is the strongest reducing agent. 2 H⁺(aq) + 2 e^- → H₂(g) ; 0.00 V K⁺(aq) + e^- → K(s) ; -2.93 V F₂(g) + 2 e^- → 2 F^-(aq) ; 2.87 V Al³⁺(aq) + 3 e^- → Al(s) ; -1.66 V Pb²⁺(aq) + 2 e^- → Pb(s) ; -0.13 V

Which of the following species are likely to behave as oxidizing agents? Li(s) ,H₂(g) ,MnO₄^-(aq) ,and Cl^-(aq)

The unit for electromotive force,emf,is the volt.1 volt is equal to _____.

According to the cell notation below,which of the following species is undergoing reduction? Ni | Ni²⁺(aq) || Mn²⁺(aq) | MnO₂(s) | Pt(s)

Use the following standard reduction potentials to determine which species is the strongest oxidizing agent. Fe²⁺(aq) + 2 e^- → Fe(s) ; E° = -0.41 V Pt²⁺(aq) + 2 e^- → Pt(s) ; E° = 1.18 V Cr₂O₇^2-(aq) + 14 H⁺(aq) + 6 e^- → 2 Cr³⁺(aq) + 7 H₂O( ) ; E° = 1.33 V

Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt²⁺(aq) + 2 e− → Pt(s) ; E° = 1.180 V Pb²⁺(aq) + 2 e− → Pb(s) ; E° = -0.130 V

Use the following standard reduction potentials to determine which species is the best oxidizing agent. O₂(g) + 4 H⁺(aq) + 4 e^- → 2 H₂O( ) E° = +1.229 V Hg₂²⁺(aq) + 2 e^- → 2 Hg( ) E° = +0.789 V I₂(s) + 2 e^- → 2 I^-(aq) E° = +0.535 V

Which of the following is the cell notation for a voltaic cell based on the following reaction? Cu²⁺(aq) + Pb(s) + SO₄^2-(aq) → Cu(s) + PbSO₄(s)

Write a balanced net ionic equation for the overall reaction represented by the following cell notation. Cu(s) | Cu²⁺(aq) || Mn²⁺(aq) | Mn(s)

Which of the following is the cell notation for a cell in which the hydrogen electrode is the anode and the cathode half-reaction is Co³⁺(aq) + e− → Co²⁺(aq) ?

Which of the following overall chemical equations is responsible for generating electricity in fuel cells used in NASA's Space Shuttle programs?

Quiz 20: Principles of Chemical Reactivity: Electron Transfer Reactions

Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb2+(aq) + 2 e- → Pb(s) E° = -0.126 V Fe3+(aq) + e- → Fe2+(s) E° = +0.771 V

Write a balanced net ionic equation for the overall reaction represented by the following cell notation. Al(s) | Al3+(aq) || Cl2(g) | Cl-(aq) | Pt(s)

An SHE electrode has been assigned a standard reduction potential,E°,of 0.00 volts.Which of the following reactions will occur at this electrode?

Calculate the standard cell potential ( ) for the reaction 2 Ag(s) + Co2+(aq) → 2 Ag+(aq) + Co(s) . The standard reduction potentials are as follows: Ag+(aq) + e−→ Ag(s) E° = 0.8 V Co2+(aq) +2 e−→ Co(s) E° = -0.277 V

In the given electrochemical cell,which of the following is the cathode half-reaction? Zn(s) | Zn2+(aq) || Fe3+(aq) ,Fe2+(aq) | Pt(s)

Calculate E°cell for the cell for the reaction 2 Ga(s) + 3 Sn4+(aq) → 3 Sn2+(aq) +2 Ga3+(aq) . The standard reduction potentials are as follows: Ga3+(aq) + 3 e− → Ga(s) E° = -0.55 V Sn4+(aq) + 2 e− → Sn2+(aq) E° = 0.15V ​

Use the following standard reduction potentials to determine which species is the strongest reducing agent. 2 H+(aq) + 2 e- → H2(g) ; 0.00 V K+(aq) + e- → K(s) ; -2.93 V F2(g) + 2 e- → 2 F-(aq) ; 2.87 V Al3+(aq) + 3 e- → Al(s) ; -1.66 V Pb2+(aq) + 2 e- → Pb(s) ; -0.13 V

Which of the following species are likely to behave as oxidizing agents? Li(s) ,H2(g) ,MnO4-(aq) ,and Cl-(aq)

The unit for electromotive force,emf,is the volt.1 volt is equal to _____.

According to the cell notation below,which of the following species is undergoing reduction? Ni | Ni2+(aq) || Mn2+(aq) | MnO2(s) | Pt(s)

Use the following standard reduction potentials to determine which species is the strongest oxidizing agent. Fe2+(aq) + 2 e- → Fe(s) ; E° = -0.41 V Pt2+(aq) + 2 e- → Pt(s) ; E° = 1.18 V Cr2O72-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O( ) ; E° = 1.33 V

Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt2+(aq) + 2 e− → Pt(s) ; E° = 1.180 V Pb2+(aq) + 2 e− → Pb(s) ; E° = -0.130 V

Consider the following half-reactions. Ag+(aq) + e- → Ag(s) E° = +0.80 V Cu2+(aq) + 2 e- → Cu(s) E° = +0.34 V Pb2+(aq) + 2 e- → Pb(s) E° = -0.13 V Fe2+(aq) + 2 e- → Fe(s) E° = -0.44 V Al3+(aq) + 3 e- → Al(s) E° = -1.66 V Which of the following species will oxidize lead,Pb(s) ?

Use the following standard reduction potentials to determine which species is the best oxidizing agent. O2(g) + 4 H+(aq) + 4 e- → 2 H2O( ) E° = +1.229 V Hg22+(aq) + 2 e- → 2 Hg( ) E° = +0.789 V I2(s) + 2 e- → 2 I-(aq) E° = +0.535 V

Consider the following half-reactions. Cl2(g) + 2 e- → 2 Cl-(aq) E° = +1.36 V Ag+(aq) + e- → Ag(s) E° = +0.80 V Cu2+(aq) + 2 e- → Cu(s) E° = +0.34 V Sn2+(aq) + 2 e- → Sn(s) E° = -0.14 V Al3+(aq) + 3 e- → Al(s) E° = -1.66 V Which of the following species will reduce Cu2+(aq) ion?

Which of the following is the cell notation for a voltaic cell based on the following reaction? Cu2+(aq) + Pb(s) + SO42-(aq) → Cu(s) + PbSO4(s)

Write a balanced net ionic equation for the overall reaction represented by the following cell notation. Cu(s) | Cu2+(aq) || Mn2+(aq) | Mn(s)

Which of the following is the cell notation for a cell in which the hydrogen electrode is the anode and the cathode half-reaction is Co3+(aq) + e− → Co2+(aq) ?