Question 1

Given that E° = +0.897 V,calculate E at 25°C for Pb(s)∣ Pb²⁺(0.0400 M)∣∣ Fe³⁺(0.200 M),Fe²⁺(0.0100 M)∣ Pt(s)

Accepted Answer

The answer of Given that E° = +0.897 V,calculate E...

Question 2

Which battery does not use MnO₂(s)as a cell reactant?

Accepted Answer

A "ni-cad" battery, or nickel-cadmium battery, uses nickel oxide hydroxide and metallic cadmium as electrodes, not manganese dioxide.

Question 3

The following cell has a potential of 0.45 V at 25°C. Pt(s)∣ H₂(1 atm)|H⁺(? M)∣∣ Cl^-(1 M)∣ Hg₂Cl₂(s)|Hg(l) The standard half-cell potential for the half-reaction Hg₂Cl₂(s)+ 2 e^- → 2 Hg(l)+ 2 Cl^-(aq)is 0.28 V.What is the pH in the anode compartment?

Accepted Answer

The Nernst equation can be used to calculate the cell potential at non-standard conditions: Ecell = E°cell - (RT/nF)lnQ where E°cell is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred in the reaction, F is Faraday's constant, and Q is the reaction quotient. In this case, the reaction is: Hg₂Cl₂(s) + 2 e^- → 2 Hg(l) + 2 Cl^-(aq) The reaction quotient can be expressed as: Q = [Hg2+][Cl-]^2 / [HgS][Cl-] = [Hg2+][Cl-] / [HgS] Since the solid HgS does not appear in the reaction quotient expression, its activity is taken to be 1. Therefore, the reaction quotient simplifies to: Q = [Hg2+][Cl-]^2 At equilibrium, the cell potential is equal to the standard cell potential. Therefore: 0.45 V = 0.28 V - (0.0257 V/K)(log([Hg2+][Cl-]^2) - log(1)) Simplifying: log([Hg2+][Cl-]^2) = 4.08 [Hg2+][Cl-]^2 = 10^4.08 [Hg2+] = sqrt(10^4.08 / [Cl-]^2) = 704.4 M Now we need to find the pH of the anode compartment, which is where the HgS and Hg2+ are located. We can use the following half-reaction to find the relationship between [Hg2+] and pH: 2 Hg2+ + 2 e^- + 2 H2O⁺ → Hg22+(aq) + 4 HSO4-^- + 4 H+ The standard reduction potential for this half-reaction is 1.23 V, which is sufficiently far from the potential of the cell in order to assume that this reaction is at equilibrium. Therefore, the Nernst equation can be used to find the relationship between [Hg2+] and pH: E° = 1.23 V E = E° - (RT/nF)ln[Hg2+][H2O]^2 / [HSO4-]^4 0.45 V = 1.23 V - (0.0257 V/K)(ln[Hg2+] + 2ln[H2O] - 4ln[HSO4-]) ln[Hg2+] = 28.53 - 44.50pH [Hg2+] = e^(28.53 - 44.50pH) Substituting the value for [Hg2+] obtained earlier: 704.4 M = e^(28.53 - 44.50pH) ln 704.4 = 28.53 - 44.50pH pH = (28.53 - ln 704.4) / 44.50 = 2.9

Question 4

A cell based on the reaction below has a standard potential of +0.42 V at 25°C.If all of the species are at standard conditions except [H⁺],at what pH will the cell have a potential of zero? H₂O₂(aq)+ 2 H⁺(aq)+ 2 Cl^-(aq)→ Cl₂(aq)+ 2 H₂O(l)

Accepted Answer

To find the pH at which the cell potential is zero, we need to use the Nernst equation: E = E° - (RT/nF) ln(Q) where E is the cell potential, E° is the standard potential, R is the gas constant, T is the temperature, n is the number of electrons transferred in the reaction, F is the Faraday constant, and Q is the reaction quotient. At equilibrium, the reaction quotient Q is equal to the equilibrium constant K: Q = K We can write the reaction quotient Q in terms of the concentrations of the species involved in the reaction: Q = [Cl₂][H⁺]^2/[H₂O][H⁺]^2 Simplifying, we get: Q = [Cl₂]/[H₂O] At the point where the cell potential is zero, E = 0, so we can solve for the pH using the Nernst equation: 0 = 0.42 - (0.0257/2) ln([Cl₂]/[H₂O]) ln([Cl₂]/[H₂O]) = 32.74 [Cl₂]/[H₂O] = e^(32.74) = 2.514 x 10^14 We know that [H⁺] is the only concentration that is not at standard conditions. Let's call its concentration x. Then, we can express the other concentrations in terms of x: [H⁺] = x [H₂O] = [H⁺] = x [Cl₂] = Q[H₂O] = (2.514 x 10^14) x Substituting these expressions into the equation for the reaction quotient, we get: K = [Cl₂]/[H₂O] = (2.514 x 10^14) x/x = 2.514 x 10^14 The equilibrium constant K is related to the standard free energy change ΔG° by: ΔG° = -RT ln(K) Substituting the values we know, we get: ΔG° = -(8.314 J/mol*K)(298 K) ln(2.514 x 10^14) ΔG° = -578.3 kJ/mol We also know that: ΔG° = -nFE° where n is the number of electrons transferred (2 in this case) and F is the Faraday constant. Substituting the values we know, we get: -578.3 kJ/mol = -2F(96500 C/mol)(0.42 V) F = 96485 C/mol Now, we can use the Henderson-Hasselbalch equation to find the pH at which [H⁺] = x: pH = pKa + log([A-]/[HA]) In this case, H⁺ is the acid and H₂O is the conjugate base. We can find the pKa from the standard free energy change: ΔG° = -RT ln(K) K = e^(-ΔG°/RT) K = [A-][H+]/[HA] pKa = -log(Ka) = -log([H+][A-]/[HA]) = -log([H+]) - log([A-]/[HA]) = -log([H+]) - log(Q) pKa = -log([H+]) - 32.74 At the point where the cell potential is zero, K = 2.514 x 10^14, so: pKa = -log([H+]) - log(2.514 x 10^14) The acid and conjugate base are both equal to x, so we can simplify the Henderson-Hasselbalch equation to: pH = pKa + log(1) = pKa pH = -log([H+]) - 32.74 -pH = log([H+]) + 32.74 [H+] = 10^(-pH-32.74) [H+] = 10^(-7.09) Therefore, the pH at which the cell potential is zero is 7.09. The answer is B.

Question 5

Consider the following cell: Pt(s)∣ H₂(g,p₁)∣ H⁺(aq,pH_A)∣∣ H⁺(aq,pH_C)∣ H₂(g,p₂)∣ Pt(s) Where pH_A is the pH of the aqueous solution in the anode half-cell and pH_C is the pH of the aqueous solution in the cathode half-cell.If the partial pressure of H₂(g)is the same for both half-cells, (p₁ = p₂),then E for the cell at 25°C is

Accepted Answer

The answer of Consider the following cell: Pt(s)∣ H₂(g,p₁)∣ H⁺(aq,pH_A)∣∣ H⁺(aq,pH_C)∣...

Question 6

The cell reaction for a lead storage battery is: Pb(s)+ PbO₂(s)+ 2 H⁺(aq)+ 2 HSO₄^-(aq)→ 2 PbSO₄(s)+ 2 H₂O(l) E° = +1.92 V To provide a potential of about 12 V,one could

Accepted Answer

To obtain a greater potential, the cells can be connected in series. In this case, 6 cells can be connected to get a total potential of 6 x 2.1 V = 12.6 V. pH adjustment or increase in surface area of Pb and PbO2 electrodes will not affect the potential of the cell.

Question 7

Given that E°_red = -0.26 V for Ni²⁺/Ni at 25°C,find E° and E for the concentration cell expressed using shorthand notation below. Ni(s)∣ Ni²⁺(aq,1.0 × 10^-5 M)∣∣ Ni²⁺(aq,0.100 M)∣ Ni(s)

Accepted Answer

The shorthand notation provided in the problem is as follows: Ni(s) ∣ Ni²⁺(aq, 1.0 × 10^-5 M) ∣∣ Ni²⁺(aq, 0.100 M) ∣ Ni(s) To find E°, we can use the given standard reduction potential: E°_red = -0.26 V E° = E°(reduction at cathode) - E°(reduction at anode) = 0 - (-0.26 V) = +0.26 V To find E, we can use the equation: E = E° - (0.0592/n) log(Q) where n is the number of electrons transferred in the reaction and Q is the reaction quotient. In this case, the cell reaction is: Ni²⁺ + 2e- → Ni + 2^2- So n = 2. At equilibrium, the reaction quotient Q is: Q = [Ni2+(aq)]/[S2-(aq)]2 Since the concentrations of Ni2+ and S2- are the same on both sides of the cell, Q = 1. So, E = E° - (0.0592/n) log(Q) = E° - (0.0592/2) log(1) = E° = +0.26 V Therefore, the final answer is: Choice B) E° = 0.00 V and E = +0.12 V, as E° = +0.26 V and E = E° = +0.26 V - (0.0592/2) log(1) = +0.12 V.

Question 8

Calculate the equilibrium constant,K,at 25°C for the galvanic cell reaction shown below:

Accepted Answer

The answer of Calculate the equilibrium constant,K,at 25°C for the...

Question 9

A particular 12V battery is based on a reaction having a standard cell potential,E° = +1.92 V.What happens when the battery "dies"?

Accepted Answer

The standard cell potential (E°) of the battery is +1.92V, which means under standard conditions, the reaction in the battery can produce a potential difference of 1.92V. When the battery "dies", it means that the reaction has reached equilibrium, and there is no potential difference between the anode and cathode. Therefore, the cell potential (E) would be 0V, while the standard cell potential (E°) remains the same at +1.92V. So, the correct answer is C) E° = +1.92 V and E = 0 V.

Question 10

Given p_H2 = 0.100 atm,[Cd²⁺] = 0.200 M,and [H⁺] = 1.00 × 10^-5 M,calculate E at 25°C for a cell based on the reaction: Cd(s)+ 2 H⁺(aq) $\rarr$H₂(g)+ Cd²⁺(aq)E^o = +0.40 V.

Accepted Answer

First, we need to calculate the concentration of [HS-], which is:

[HS-] = [HS1U1P1+S1S1P0] - [H+]

[H+] = 10^-pH 
pH = -log[H+]

Since this is an acidic solution, we can assume that [H+] ≈ [S1S1P0] ≈ 10^-5 M

[HS-] = (1.00 × 10^-5 M) - (10^-5 M)
[HS-] = 0.00 M

Next, we need to calculate the concentration of [Cd2+] which is:

[Ksp] = [Cd2+][S1U1B12S1U1B0]

Since [CdS1U1P12S1S1P0] is not involved in the reaction, we can ignore its concentration.

Therefore,

[Cd2+] = [S1U1B12S1U1B0] = Ksp/[Cd2+] = (0.100 atm)/(1.33 × 10^-23 atm^2) = 7.52 × 10^-1 M

Now we can calculate the cell potential:

Ecell = E°cell - (RT/nF)lnQ

where E°cell = +0.40 V (given), R = 8.314 J/mol*K, T = 298 K, n = 2 (from the balanced equation), F = 9.649 × 10^4 C/mol, and Q = [(HS-)^2][Cd2+]/(1M)^2 = [HS-]^2[Cd2+]

Ecell = +0.40 V - (8.314 J/mol*K)(298 K)/(2)(9.649 × 10^4 C/mol)ln([HS-]^2[Cd2+]/(1M)^2)

Ecell = +0.15 V

Therefore, the answer is C) +0.15 V.

Question 11

The equilibrium constant,K,can be calculated from

Accepted Answer

The equilibrium constant, K, can be calculated from the standard electrode potential (E°), not from the electrode potential (E) under non-standard conditions. The relationship between E° and K is given by the Nernst equation.

Question 12

At 25°C,E° = +1.88 V for a cell based on the reaction 3 AgCl(s)+ Al(s)→ 3 Ag(s)+ Al³⁺(aq)+ 3 Cl^-(aq). Find the cell potential E if [Al³⁺] = 0.20 M and [Cl^-] = 0.010 M.

Accepted Answer

The Nernst equation can be used to calculate the cell potential with non-standard concentration of reactants and products. The equation is:

E = E° - (RT/nF) ln(Q)

where E° is the standard cell potential, R is the gas constant, T is the temperature in kelvin, n is the number of electrons transferred in the reaction, F is Faraday's constant, and Q is the reaction quotient.

For the given reaction, n = 3 because three electrons are transferred. Using the given values, we get:

E = 1.88 - (0.0257/3) ln((0.010)^3/(0.20)^3) = 2.01 V

Therefore, the correct answer is A, +2.01 V.

Question 13

For a particular cell based on the reaction: 3 AgCl(s)+ Al(s)→ 3 Ag(s)+ Al³⁺(aq)+ 3 Cl^-(aq) E = +1.750 V and E° = +1.884 V at 25°C. What is the value of the equilibrium constant,K,at 25°C for the reaction?

Accepted Answer

The answer of For a particular cell based on the...

Question 14

Which is most often used in the laboratory to measure pH?

Accepted Answer

A glass electrode is the most commonly used instrument for measuring pH in a laboratory setting. It consists of a thin glass membrane that is sensitive to hydrogen ions and generates an electrical signal that is proportional to the pH of the solution being tested. The other options, such as a standard hydrogen electrode, Daniell cell, and conductivity cell, are not typically used for pH measurement.

Question 15

Consider the half-reaction: MnO₄^- (aq)+ 8 H⁺ (aq)+ 5 e^- → Mn²⁺ (aq)+ 4 H₂O(l).The formation of MnO₄^- from Mn²⁺ occurs most readily when the solution is

Accepted Answer

The answer of Consider the half-reaction: MnO₄^- (aq)+ 8 H⁺...

Question 16

When a cell reaction reaches equilibrium,

Accepted Answer

At equilibrium, the cell potential (E) becomes zero because there is no net movement of electrons. The standard cell potential (E°) is a constant value that depends on the nature of the reactants and products at standard conditions and does not change when equilibrium is reached.

Question 17

Ag⁺(aq)+ e^- → Ag(s)E° = +0.800 V AgBr(s)+ e^- → Ag(s)+ Br^-(aq)E° = +0.071 V Br₂(l)+ 2 e^- → 2 Br^-(aq)E° = +1.066 V Use some of the data above to calculate K_sp at 25°C for AgBr.

Accepted Answer

The Ksp for AgBr can be calculated using the Nernst equation and the given standard reduction potentials. The difference in potentials for the dissolution of AgBr is 0.800 V - 0.071 V = 0.729 V. Using the relation ΔG° = -nFE° and ΔG° = -RTlnK, we find Ksp = 10^(-ΔG°/RT) = 10^(-0.729/0.0592) = 4.9 × 10^-13.

Question 18

How many moles of electrons,n,are transferred in the following reduction-oxidation reaction? 2 MnO₄^-(aq)+ 16 H⁺(aq)+ 10 Cl^-(aq)→ 2 Mn²⁺(aq)+ 5 Cl₂(g)+ 8 H₂O(l)

Accepted Answer

In the given redox reaction, the total number of electrons transferred can be determined by balancing the reaction in terms of electron transfer. The Mn goes from a +7 oxidation state in MnO4- to a +2 oxidation state in Mn2+, which is a reduction involving the gain of 5 electrons per Mn atom. Since there are 2 Mn atoms involved, the total is 2 * 5 = 10 electrons. The Cl- is oxidized to Cl2, but the focus here is on the total number of electrons transferred, which is 10.

Question 19

For a particular battery based on one of the following reactions,E is expected to remain constant with time until the cell reactants are almost completely consumed.Which is the appropriate reaction?

Accepted Answer

The given reaction, 2 NiO(OH)(s)+ Cd(s)+ 2 H₂O(l)→ 2 Ni(OH)₂(s)+ Cd(OH)₂(s), involves the reaction between solid NiO(OH) and a solid Cd electrode in an acidic electrolyte. This is a typical example of an alkaline battery, where the reactants, NiO(OH) and Cd, are slowly consumed during the discharge process, and the voltage across the battery remains constant until most of the reactants are consumed. Therefore, the E is expected to remain constant with time until the cell reactants are almost completely consumed.

Question 20

If the cell reaction involves ions in solution,as the cell reaction in a galvanic cell continues,

Accepted Answer

As the cell reaction in a galvanic cell continues, the concentration of reactants decreases and the concentration of products increases, which causes a decrease in the free energy change (ΔG) and therefore a decrease in the EMF (E) for the cell. This is consistent with Le Chatelier's principle, which states that a system at equilibrium will respond to any stress imposed upon it in such a way as to counteract the stress and reestablish equilibrium. In this case, the stress is the consumption of reactants and the formation of products, which causes a shift in the direction of the reverse reaction, resulting in a decrease in E for the cell. However, the standard electrode potential (E°) for the cell remains unchanged since it is a measure of the inherent tendency for the reaction to occur under standard conditions.

Quiz 17: Electrochemistry

Given that E° = +0.897 V,calculate E at 25°C for Pb(s) ∣ Pb²⁺(0.0400 M) ∣∣ Fe³⁺(0.200 M) ,Fe²⁺(0.0100 M) ∣ Pt(s)

Which battery does not use MnO₂(s) as a cell reactant?

The following cell has a potential of 0.45 V at 25°C. Pt(s) ∣ H₂(1 atm) |H⁺(? M) ∣∣ Cl^-(1 M) ∣ Hg₂Cl₂(s) |Hg(l) The standard half-cell potential for the half-reaction Hg₂Cl₂(s) + 2 e^- → 2 Hg(l) + 2 Cl^-(aq) is 0.28 V.What is the pH in the anode compartment?

A cell based on the reaction below has a standard potential of +0.42 V at 25°C.If all of the species are at standard conditions except [H⁺],at what pH will the cell have a potential of zero? H₂O₂(aq) + 2 H⁺(aq) + 2 Cl^-(aq) → Cl₂(aq) + 2 H₂O(l)

The cell reaction for a lead storage battery is: Pb(s) + PbO₂(s) + 2 H⁺(aq) + 2 HSO₄^-(aq) → 2 PbSO₄(s) + 2 H₂O(l) E° = +1.92 V To provide a potential of about 12 V,one could

Given that E°_red = -0.26 V for Ni²⁺/Ni at 25°C,find E° and E for the concentration cell expressed using shorthand notation below. Ni(s) ∣ Ni²⁺(aq,1.0 × 10^-5 M) ∣∣ Ni²⁺(aq,0.100 M) ∣ Ni(s)

Calculate the equilibrium constant,K,at 25°C for the galvanic cell reaction shown below:

A particular 12V battery is based on a reaction having a standard cell potential,E° = +1.92 V.What happens when the battery "dies"?

Given p_H2 = 0.100 atm,[Cd²⁺] = 0.200 M,and [H⁺] = 1.00 × 10^-5 M,calculate E at 25°C for a cell based on the reaction: Cd(s) + 2 H⁺(aq) $\rarr$ H₂(g) + Cd²⁺(aq) E^o = +0.40 V.

The equilibrium constant,K,can be calculated from

At 25°C,E° = +1.88 V for a cell based on the reaction 3 AgCl(s) + Al(s) → 3 Ag(s) + Al³⁺(aq) + 3 Cl^-(aq) . Find the cell potential E if [Al³⁺] = 0.20 M and [Cl^-] = 0.010 M.

For a particular cell based on the reaction: 3 AgCl(s) + Al(s) → 3 Ag(s) + Al³⁺(aq) + 3 Cl^-(aq) E = +1.750 V and E° = +1.884 V at 25°C. What is the value of the equilibrium constant,K,at 25°C for the reaction?

Which is most often used in the laboratory to measure pH?

Consider the half-reaction: MnO₄^- (aq) + 8 H⁺ (aq) + 5 e^- → Mn²⁺ (aq) + 4 H₂O(l) .The formation of MnO₄^- from Mn²⁺ occurs most readily when the solution is

When a cell reaction reaches equilibrium,

Ag⁺(aq) + e^- → Ag(s) E° = +0.800 V AgBr(s) + e^- → Ag(s) + Br^-(aq) E° = +0.071 V Br₂(l) + 2 e^- → 2 Br^-(aq) E° = +1.066 V Use some of the data above to calculate K_sp at 25°C for AgBr.

How many moles of electrons,n,are transferred in the following reduction-oxidation reaction? 2 MnO₄^-(aq) + 16 H⁺(aq) + 10 Cl^-(aq) → 2 Mn²⁺(aq) + 5 Cl₂(g) + 8 H₂O(l)

For a particular battery based on one of the following reactions,E is expected to remain constant with time until the cell reactants are almost completely consumed.Which is the appropriate reaction?

If the cell reaction involves ions in solution,as the cell reaction in a galvanic cell continues,

Quiz 17: Electrochemistry

Given that E° = +0.897 V,calculate E at 25°C for Pb(s) ∣ Pb2+(0.0400 M) ∣∣ Fe3+(0.200 M) ,Fe2+(0.0100 M) ∣ Pt(s)

Which battery does not use MnO2(s) as a cell reactant?

The following cell has a potential of 0.45 V at 25°C. Pt(s) ∣ H2(1 atm) |H+(? M) ∣∣ Cl-(1 M) ∣ Hg2Cl2(s) |Hg(l) The standard half-cell potential for the half-reaction Hg2Cl2(s) + 2 e- → 2 Hg(l) + 2 Cl-(aq) is 0.28 V.What is the pH in the anode compartment?

A cell based on the reaction below has a standard potential of +0.42 V at 25°C.If all of the species are at standard conditions except [H+],at what pH will the cell have a potential of zero? H2O2(aq) + 2 H+(aq) + 2 Cl-(aq) → Cl2(aq) + 2 H2O(l)

The cell reaction for a lead storage battery is: Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4-(aq) → 2 PbSO4(s) + 2 H2O(l) E° = +1.92 V To provide a potential of about 12 V,one could

Given that E°red = -0.26 V for Ni2+/Ni at 25°C,find E° and E for the concentration cell expressed using shorthand notation below. Ni(s) ∣ Ni2+(aq,1.0 × 10-5 M) ∣∣ Ni2+(aq,0.100 M) ∣ Ni(s)

Calculate the equilibrium constant,K,at 25°C for the galvanic cell reaction shown below:

A particular 12V battery is based on a reaction having a standard cell potential,E° = +1.92 V.What happens when the battery "dies"?

Given pH2 = 0.100 atm,[Cd2+] = 0.200 M,and [H+] = 1.00 × 10-5 M,calculate E at 25°C for a cell based on the reaction: Cd(s) + 2 H+(aq) →\rarr→ H2(g) + Cd2+(aq) Eo = +0.40 V.

The equilibrium constant,K,can be calculated from

At 25°C,E° = +1.88 V for a cell based on the reaction 3 AgCl(s) + Al(s) → 3 Ag(s) + Al3+(aq) + 3 Cl-(aq) . Find the cell potential E if [Al3+] = 0.20 M and [Cl-] = 0.010 M.

For a particular cell based on the reaction: 3 AgCl(s) + Al(s) → 3 Ag(s) + Al3+(aq) + 3 Cl-(aq) E = +1.750 V and E° = +1.884 V at 25°C. What is the value of the equilibrium constant,K,at 25°C for the reaction?

Which is most often used in the laboratory to measure pH?

Consider the half-reaction: MnO4- (aq) + 8 H+ (aq) + 5 e- → Mn2+ (aq) + 4 H2O(l) .The formation of MnO4- from Mn2+ occurs most readily when the solution is

When a cell reaction reaches equilibrium,

Ag+(aq) + e- → Ag(s) E° = +0.800 V AgBr(s) + e- → Ag(s) + Br-(aq) E° = +0.071 V Br2(l) + 2 e- → 2 Br-(aq) E° = +1.066 V Use some of the data above to calculate Ksp at 25°C for AgBr.

How many moles of electrons,n,are transferred in the following reduction-oxidation reaction? 2 MnO4-(aq) + 16 H+(aq) + 10 Cl-(aq) → 2 Mn2+(aq) + 5 Cl2(g) + 8 H2O(l)

For a particular battery based on one of the following reactions,E is expected to remain constant with time until the cell reactants are almost completely consumed.Which is the appropriate reaction?

If the cell reaction involves ions in solution,as the cell reaction in a galvanic cell continues,

Given that E° = +0.897 V,calculate E at 25°C for Pb(s) ∣ Pb²⁺(0.0400 M) ∣∣ Fe³⁺(0.200 M) ,Fe²⁺(0.0100 M) ∣ Pt(s)

Which battery does not use MnO₂(s) as a cell reactant?

The following cell has a potential of 0.45 V at 25°C. Pt(s) ∣ H₂(1 atm) |H⁺(? M) ∣∣ Cl^-(1 M) ∣ Hg₂Cl₂(s) |Hg(l) The standard half-cell potential for the half-reaction Hg₂Cl₂(s) + 2 e^- → 2 Hg(l) + 2 Cl^-(aq) is 0.28 V.What is the pH in the anode compartment?

A cell based on the reaction below has a standard potential of +0.42 V at 25°C.If all of the species are at standard conditions except [H⁺],at what pH will the cell have a potential of zero? H₂O₂(aq) + 2 H⁺(aq) + 2 Cl^-(aq) → Cl₂(aq) + 2 H₂O(l)

The cell reaction for a lead storage battery is: Pb(s) + PbO₂(s) + 2 H⁺(aq) + 2 HSO₄^-(aq) → 2 PbSO₄(s) + 2 H₂O(l) E° = +1.92 V To provide a potential of about 12 V,one could

Given that E°_red = -0.26 V for Ni²⁺/Ni at 25°C,find E° and E for the concentration cell expressed using shorthand notation below. Ni(s) ∣ Ni²⁺(aq,1.0 × 10^-5 M) ∣∣ Ni²⁺(aq,0.100 M) ∣ Ni(s)

Given p_H2 = 0.100 atm,[Cd²⁺] = 0.200 M,and [H⁺] = 1.00 × 10^-5 M,calculate E at 25°C for a cell based on the reaction: Cd(s) + 2 H⁺(aq) $\rarr$ H₂(g) + Cd²⁺(aq) E^o = +0.40 V.

At 25°C,E° = +1.88 V for a cell based on the reaction 3 AgCl(s) + Al(s) → 3 Ag(s) + Al³⁺(aq) + 3 Cl^-(aq) . Find the cell potential E if [Al³⁺] = 0.20 M and [Cl^-] = 0.010 M.

For a particular cell based on the reaction: 3 AgCl(s) + Al(s) → 3 Ag(s) + Al³⁺(aq) + 3 Cl^-(aq) E = +1.750 V and E° = +1.884 V at 25°C. What is the value of the equilibrium constant,K,at 25°C for the reaction?

Consider the half-reaction: MnO₄^- (aq) + 8 H⁺ (aq) + 5 e^- → Mn²⁺ (aq) + 4 H₂O(l) .The formation of MnO₄^- from Mn²⁺ occurs most readily when the solution is

Ag⁺(aq) + e^- → Ag(s) E° = +0.800 V AgBr(s) + e^- → Ag(s) + Br^-(aq) E° = +0.071 V Br₂(l) + 2 e^- → 2 Br^-(aq) E° = +1.066 V Use some of the data above to calculate K_sp at 25°C for AgBr.

How many moles of electrons,n,are transferred in the following reduction-oxidation reaction? 2 MnO₄^-(aq) + 16 H⁺(aq) + 10 Cl^-(aq) → 2 Mn²⁺(aq) + 5 Cl₂(g) + 8 H₂O(l)