Question 1

For the reaction: Mg(s)+ AgNO₃(aq)→ Ag(s)+ Mg(NO₃)₂(aq) Ag⁺(aq)+ e^- → Ag(s)E° = 0.800 V Mg²⁺(aq)+ 2 e^- → Mg(s)E° = -2.356 V Estimate K_eq at 25 °C.

Accepted Answer

The equilibrium constant (K_eq) for the reaction can be calculated using the Nernst equation and the standard electrode potentials (E°) given for the half-reactions. The overall cell potential (E°_cell) is the difference between the reduction potential of silver ion (Ag^+) and the oxidation potential of magnesium (Mg), which is the negative of the given E° for Mg. Thus, E°_cell = 0.800 V - (-2.356 V) = 3.156 V. The relationship between E°_cell and K_eq at 25 °C (298 K) is given by the equation ΔG° = -nFE°_cell = -RTlnK_eq, where n is the number of moles of electrons transferred (2 in this case), F is the Faraday constant (96485 C/mol), R is the gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin. Rearranging for K_eq gives K_eq = e^(-ΔG°/(RT)) = e^((nFE°_cell)/(RT)). Substituting the values and calculating gives a K_eq of approximately 5.7 × 10^110, which matches choice A.

Question 2

Accepted Answer

Options I and IV describe the silver/silver chloride electrode and the mercury/mercurous chloride (calomel) electrode, respectively, both of which are commonly used as reference electrodes in electrochemical measurements.

Question 3

Determine the K_spat 25 °C of AgCl(s).The half-reactions are: Ag⁺(aq)+ 1 e^- → Ag(s)E° = +0.771 V AgCl(s)+ 1 e^- → Ag(s)+ Cl^-(aq) E° = +0.222 V

Accepted Answer

The Ksp can be calculated using the Nernst equation and the given standard reduction potentials. The difference in potentials (ΔE°) is 0.771 V - 0.222 V = 0.549 V. Using the relation ΔG° = -nFE° and ΔG° = -RTlnK, we find Ksp = 10^(-ΔG°/RT) = 5 × 10^-10.

Question 4

Choose the INCORRECT statement.

Accepted Answer

When E_cell < 0, the reaction is non-spontaneous (not spontaneous).

Question 5

What is the concentration of Cu²⁺(aq)in the following cell at 25 °C if the cell voltage is 1.253 V? Zn(s)∣ Zn²⁺ (aq,0.125 M) Cu²⁺(?)∣ Cu(s) Cu²⁺(aq)+ 2 e^- → Cu(s)E° = 0.340 V Zn²⁺(aq)+ 2 e^- → Zn(s)E° = -0.763 V

Accepted Answer

To solve this problem, we can use the Nernst equation: E = E° - (RT/nF)ln(Q) We know the cell voltage (E) is 1.253 V, the standard reduction potentials (E°) for the two half-reactions, and the temperature (T) is 25 °C. We can rearrange the equation to solve for the unknown concentration of Cu²⁺(aq) (Q) by plugging in the values: Q = eS1U1PnF/(RT) * (E°cell - E) = (1 mol)/(96485 C) * (0.340 V - 1.253 V) / (8.314 J/(mol*K) * 298 K) Q = 2.02 × 10^-6 The concentration of Cu²⁺(aq) is equal to the value of Q, which is approximately 1 × 10^-6. Therefore, the best answer choice is C.

Question 6

If the voltage of an electrochemical cell is negative then the cell reaction is:

Accepted Answer

A negative voltage indicates that the cell reaction is not spontaneous, meaning that it requires an input of energy in order to occur. Thus, option A is the correct choice.

Question 7

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

Accepted Answer

E°cell = E°cathode - E°anode. Here, Al is reduced (E° = -1.676 V) and Zn is oxidized (E° = -0.763 V). Thus, E°cell = (-0.763) - (-1.676) = 0.913 V.

Question 8

For the reaction: Mg(s)+ AgNO₃(aq)→ Ag(s)+ Mg(NO₃)₂(aq) Ag⁺(aq)+ e^- → Ag(s)E° = 0.800 V Mg²⁺(aq)+ 2 e^- → Mg(s)E° = -2.356 V Is the reaction spontaneous and why?

Accepted Answer

The reaction is spontaneous because the overall standard electrode potential (E°) for the reaction is positive. This is determined by calculating the difference between the reduction potential of silver ion (Ag+) and the oxidation potential of magnesium (Mg), which is the negative of its given reduction potential. The positive E° value indicates spontaneity.

Question 9

Glass electrodes are used for direct measurement of:

Accepted Answer

Glass electrodes are commonly used for the direct measurement of pH, as they are able to detect the concentration of hydrogen ions in a solution.

Question 10

The lithium-ion battery is based on:

Accepted Answer

The answer of The lithium-ion battery is based on:
...

Question 11

Choose the INCORRECT statement.

Accepted Answer

Explanation: One Faraday of charge will plate out one mole of Cu(s) from a solution of CuSO4(aq), not two moles. This is because copper (Cu) has a valency of +2, and it requires two moles of electrons (2 Faradays) to reduce one mole of Cu2+ ions to Cu(s).

Question 12

Choose the INCORRECT statement.

Accepted Answer

Primary batteries cannot be recharged by passing electricity through the battery.

Question 13

Consider the cell: Ni(s)∣ Ni²⁺ (aq,?M) Cu²⁺ (aq,0.136 M)∣ Cu(s) Ni²⁺(aq)/Ni(s) E° = -0.257 V Cu²⁺(aq)/Cu(s) E° = 0.340 V The measured potential of the cell is 0.621 V.What is [Ni^2+] at 25 °C?

Accepted Answer

The answer of Consider the cell: Ni(s)∣ Ni²⁺ (aq,?M) Cu²⁺...

Question 14

Choose the INCORRECT statement.

Accepted Answer

The incorrect statement is E because the active metal used in cathodic protection is actually called a sacrificial anode, not a sacrificial cathode.

Question 15

What is the cell diagram for the spontaneous cell involving the Fe³⁺(aq)∣ Fe²⁺(aq)∣ Pt(s)(0.771V)half cell and the Zn²⁺(aq)∣ Zn(s)(-0.763 V)half cell?

Accepted Answer

The answer of What is the cell diagram for the...

Question 16

Determine E°_cell for the reaction: 2 Ag⁺(aq)+ Mg(s)→ 2 Ag(s)+ Mg²⁺(aq).The half reactions are: Mg²⁺(aq)+ 2 e^- → Mg(s)E° = -2.356 V Ag⁺(aq)+ e^- → Ag(s)E° = 0.800 V

Accepted Answer

The answer of Determine E°_cell for the reaction: 2 Ag⁺(aq)+...

Question 17

What is the concentration of Al³⁺(aq)in the following cell at 25 °C if the cell voltage is 1.486 V? Al(s)∣ Al³⁺(aq,?) Sn²⁺(aq,3.21 × 10^-4 M)∣ Sn(s) Al³⁺(aq)+ 3 e^- → Al(s)E° = -1.676 V Sn²⁺(aq)+ 2 e^- → Sn(s)E° = -0.137 V

Accepted Answer

The answer of What is the concentration of Al³⁺(aq)in the...

Question 18

Calculate the K_sp of lead iodide from the following standard electrode potentials,at 25 °C: 2 e^- + PbI₂(S)→ Pb(s)+ 2 I^-(aq)E° = -0.365 V 2 e^- + Pb²⁺(aq)→ Pb(s)E° = -0.126 V

Accepted Answer

The correct Ksp value can be calculated using the Nernst equation and the given standard electrode potentials.

Question 19

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

Accepted Answer

The E°cell can be calculated using the formula E°cell = E°cathode - E°anode. 
In this case, since Pb is reduced (gains electrons) and Zn is oxidized (loses electrons), the half reaction with the more positive E° value (Zn) is the anode and the one with the more negative E° value (Pb) is the cathode. 
Therefore, E°cell = E°cathode - E°anode = -0.125 V - (-0.763 V) = 0.638 V. 
Note that the value is positive, indicating that the reaction is spontaneous in the forward direction (as written).

Question 20

In the cell Pt(s)∣ Fe²⁺(aq)∣ Fe³⁺(aq) Cu²⁺(aq)∣ Cu(s)which will increase the cell voltage the most?

Accepted Answer

The cell potential is determined by the difference in electrode potentials for the oxidation and reduction half-reactions. In this case, the two half-reactions are:
Pt(s) + 2 FeS1U1P12+(aq) + 2 e- → 2 FeS1U1P13+(aq)          (oxidation at anode)
CuS1U1P12+(aq) + e- → Cu(s) + S1S1P0(aq)                            (reduction at cathode)

To increase the cell potential, we need to increase the difference in electrode potentials. We can do this by either increasing the reduction potential (for the reduction half-reaction) or decreasing the oxidation potential (for the oxidation half-reaction).

Of the options given, doubling the concentration of FeS1U1P12+S1S1P0 will have the greatest effect on the oxidation half-reaction, as it will increase the concentration of the oxidizing agent (FeS1U1P12+). This will shift the equilibrium towards the reduction half-reaction, increasing the cell potential.

Therefore, the best choice is D (double [FeS1U1P12+S1S1P0]).

Quiz 19: Electrochemistry

For the reaction: Mg(s) + AgNO₃(aq) → Ag(s) + Mg(NO₃) ₂(aq) Ag⁺(aq) + e^- → Ag(s) E° = 0.800 V Mg²⁺(aq) + 2 e^- → Mg(s) E° = -2.356 V Estimate K_eq at 25 °C.

Which of the following are commonly used as alternate standard electrodes: I.Ag(s) | AgCl(s) | Cl^- ((aq,1.0 M) II.Hg(l) | HgCl₂(s) | Cl^- (aq,1.0 M) III.Ag(s) | AgCl(s) | Cl^- ((aq,1.0 M) IV.Hg(l) | Hg₂Cl₂(s) | Cl^- (aq,1.0 M) V.Hg(s) | Hg₂Cl₂(s) | Cl^- (aq,1.0 M)

Determine the K_spat 25 °C of AgCl(s) .The half-reactions are: Ag⁺(aq) + 1 e^- → Ag(s) E° = +0.771 V AgCl(s) + 1 e^- → Ag(s) + Cl^-(aq) E° = +0.222 V

Choose the INCORRECT statement.

What is the concentration of Cu²⁺(aq) in the following cell at 25 °C if the cell voltage is 1.253 V? Zn(s) ∣ Zn²⁺ (aq,0.125 M) Cu²⁺(?) ∣ Cu(s) Cu²⁺(aq) + 2 e^- → Cu(s) E° = 0.340 V Zn²⁺(aq) + 2 e^- → Zn(s) E° = -0.763 V

If the voltage of an electrochemical cell is negative then the cell reaction is:

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

For the reaction: Mg(s) + AgNO₃(aq) → Ag(s) + Mg(NO₃) ₂(aq) Ag⁺(aq) + e^- → Ag(s) E° = 0.800 V Mg²⁺(aq) + 2 e^- → Mg(s) E° = -2.356 V Is the reaction spontaneous and why?

Glass electrodes are used for direct measurement of:

The lithium-ion battery is based on:

Choose the INCORRECT statement.

Choose the INCORRECT statement.

Consider the cell: Ni(s) ∣ Ni²⁺ (aq,?M) Cu²⁺ (aq,0.136 M) ∣ Cu(s) Ni²⁺(aq) /Ni(s) E° = -0.257 V Cu²⁺(aq) /Cu(s) E° = 0.340 V The measured potential of the cell is 0.621 V.What is [Ni^2+] at 25 °C?

Choose the INCORRECT statement.

What is the cell diagram for the spontaneous cell involving the Fe³⁺(aq) ∣ Fe²⁺(aq) ∣ Pt(s) (0.771V) half cell and the Zn²⁺(aq) ∣ Zn(s) (-0.763 V) half cell?

Determine E°_cell for the reaction: 2 Ag⁺(aq) + Mg(s) → 2 Ag(s) + Mg²⁺(aq) .The half reactions are: Mg²⁺(aq) + 2 e^- → Mg(s) E° = -2.356 V Ag⁺(aq) + e^- → Ag(s) E° = 0.800 V

What is the concentration of Al³⁺(aq) in the following cell at 25 °C if the cell voltage is 1.486 V? Al(s) ∣ Al³⁺(aq,?) Sn²⁺(aq,3.21 × 10^-4 M) ∣ Sn(s) Al³⁺(aq) + 3 e^- → Al(s) E° = -1.676 V Sn²⁺(aq) + 2 e^- → Sn(s) E° = -0.137 V

Calculate the K_sp of lead iodide from the following standard electrode potentials,at 25 °C: 2 e^- + PbI₂(S) → Pb(s) + 2 I^-(aq) E° = -0.365 V 2 e^- + Pb²⁺(aq) → Pb(s) E° = -0.126 V

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

In the cell Pt(s) ∣ Fe²⁺(aq) ∣ Fe³⁺(aq) Cu²⁺(aq) ∣ Cu(s) which will increase the cell voltage the most?

Quiz 19: Electrochemistry

For the reaction: Mg(s) + AgNO3(aq) → Ag(s) + Mg(NO3) 2(aq) Ag+(aq) + e- → Ag(s) E° = 0.800 V Mg2+(aq) + 2 e- → Mg(s) E° = -2.356 V Estimate Keq at 25 °C.

Which of the following are commonly used as alternate standard electrodes: I.Ag(s) | AgCl(s) | Cl- ((aq,1.0 M) II.Hg(l) | HgCl2(s) | Cl- (aq,1.0 M) III.Ag(s) | AgCl(s) | Cl- ((aq,1.0 M) IV.Hg(l) | Hg2Cl2(s) | Cl- (aq,1.0 M) V.Hg(s) | Hg2Cl2(s) | Cl- (aq,1.0 M)

Determine the Ksp at 25 °C of AgCl(s) .The half-reactions are: Ag+(aq) + 1 e- → Ag(s) E° = +0.771 V AgCl(s) + 1 e- → Ag(s) + Cl-(aq) E° = +0.222 V

Choose the INCORRECT statement.

What is the concentration of Cu2+(aq) in the following cell at 25 °C if the cell voltage is 1.253 V? Zn(s) ∣ Zn2+ (aq,0.125 M) Cu2+(?) ∣ Cu(s) Cu2+(aq) + 2 e- → Cu(s) E° = 0.340 V Zn2+(aq) + 2 e- → Zn(s) E° = -0.763 V

If the voltage of an electrochemical cell is negative then the cell reaction is:

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

For the reaction: Mg(s) + AgNO3(aq) → Ag(s) + Mg(NO3) 2(aq) Ag+(aq) + e- → Ag(s) E° = 0.800 V Mg2+(aq) + 2 e- → Mg(s) E° = -2.356 V Is the reaction spontaneous and why?

Glass electrodes are used for direct measurement of:

The lithium-ion battery is based on:

Choose the INCORRECT statement.

Choose the INCORRECT statement.

Consider the cell: Ni(s) ∣ Ni2+ (aq,?M) Cu2+ (aq,0.136 M) ∣ Cu(s) Ni2+(aq) /Ni(s) E° = -0.257 V Cu2+(aq) /Cu(s) E° = 0.340 V The measured potential of the cell is 0.621 V.What is [Ni2+] at 25 °C?

Choose the INCORRECT statement.

What is the cell diagram for the spontaneous cell involving the Fe3+(aq) ∣ Fe2+(aq) ∣ Pt(s) (0.771V) half cell and the Zn2+(aq) ∣ Zn(s) (-0.763 V) half cell?

Determine E°cell for the reaction: 2 Ag+(aq) + Mg(s) → 2 Ag(s) + Mg2+(aq) .The half reactions are: Mg2+(aq) + 2 e- → Mg(s) E° = -2.356 V Ag+(aq) + e- → Ag(s) E° = 0.800 V

What is the concentration of Al3+(aq) in the following cell at 25 °C if the cell voltage is 1.486 V? Al(s) ∣ Al3+(aq,?) Sn2+(aq,3.21 × 10-4 M) ∣ Sn(s) Al3+(aq) + 3 e- → Al(s) E° = -1.676 V Sn2+(aq) + 2 e- → Sn(s) E° = -0.137 V

Calculate the Ksp of lead iodide from the following standard electrode potentials,at 25 °C: 2 e- + PbI2(S) → Pb(s) + 2 I-(aq) E° = -0.365 V 2 e- + Pb2+(aq) → Pb(s) E° = -0.126 V

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

In the cell Pt(s) ∣ Fe2+(aq) ∣ Fe3+(aq) Cu2+(aq) ∣ Cu(s) which will increase the cell voltage the most?

For the reaction: Mg(s) + AgNO₃(aq) → Ag(s) + Mg(NO₃) ₂(aq) Ag⁺(aq) + e^- → Ag(s) E° = 0.800 V Mg²⁺(aq) + 2 e^- → Mg(s) E° = -2.356 V Estimate K_eq at 25 °C.

Which of the following are commonly used as alternate standard electrodes: I.Ag(s) | AgCl(s) | Cl^- ((aq,1.0 M) II.Hg(l) | HgCl₂(s) | Cl^- (aq,1.0 M) III.Ag(s) | AgCl(s) | Cl^- ((aq,1.0 M) IV.Hg(l) | Hg₂Cl₂(s) | Cl^- (aq,1.0 M) V.Hg(s) | Hg₂Cl₂(s) | Cl^- (aq,1.0 M)

Determine the K_spat 25 °C of AgCl(s) .The half-reactions are: Ag⁺(aq) + 1 e^- → Ag(s) E° = +0.771 V AgCl(s) + 1 e^- → Ag(s) + Cl^-(aq) E° = +0.222 V

What is the concentration of Cu²⁺(aq) in the following cell at 25 °C if the cell voltage is 1.253 V? Zn(s) ∣ Zn²⁺ (aq,0.125 M) Cu²⁺(?) ∣ Cu(s) Cu²⁺(aq) + 2 e^- → Cu(s) E° = 0.340 V Zn²⁺(aq) + 2 e^- → Zn(s) E° = -0.763 V

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

For the reaction: Mg(s) + AgNO₃(aq) → Ag(s) + Mg(NO₃) ₂(aq) Ag⁺(aq) + e^- → Ag(s) E° = 0.800 V Mg²⁺(aq) + 2 e^- → Mg(s) E° = -2.356 V Is the reaction spontaneous and why?

Consider the cell: Ni(s) ∣ Ni²⁺ (aq,?M) Cu²⁺ (aq,0.136 M) ∣ Cu(s) Ni²⁺(aq) /Ni(s) E° = -0.257 V Cu²⁺(aq) /Cu(s) E° = 0.340 V The measured potential of the cell is 0.621 V.What is [Ni^2+] at 25 °C?

What is the cell diagram for the spontaneous cell involving the Fe³⁺(aq) ∣ Fe²⁺(aq) ∣ Pt(s) (0.771V) half cell and the Zn²⁺(aq) ∣ Zn(s) (-0.763 V) half cell?

Determine E°_cell for the reaction: 2 Ag⁺(aq) + Mg(s) → 2 Ag(s) + Mg²⁺(aq) .The half reactions are: Mg²⁺(aq) + 2 e^- → Mg(s) E° = -2.356 V Ag⁺(aq) + e^- → Ag(s) E° = 0.800 V

What is the concentration of Al³⁺(aq) in the following cell at 25 °C if the cell voltage is 1.486 V? Al(s) ∣ Al³⁺(aq,?) Sn²⁺(aq,3.21 × 10^-4 M) ∣ Sn(s) Al³⁺(aq) + 3 e^- → Al(s) E° = -1.676 V Sn²⁺(aq) + 2 e^- → Sn(s) E° = -0.137 V

Calculate the K_sp of lead iodide from the following standard electrode potentials,at 25 °C: 2 e^- + PbI₂(S) → Pb(s) + 2 I^-(aq) E° = -0.365 V 2 e^- + Pb²⁺(aq) → Pb(s) E° = -0.126 V

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

In the cell Pt(s) ∣ Fe²⁺(aq) ∣ Fe³⁺(aq) Cu²⁺(aq) ∣ Cu(s) which will increase the cell voltage the most?