Question 1

What is the value of D G ° _rxn for the following unbalanced reaction? Mg (s) + Au³⁺ (aq)→Mg²⁺ (aq) + Au (s) E ° _red(Mg²⁺/Mg) = - 2.37 V and E ° _red(Au³⁺/Au) = 1.50 V

Accepted Answer

The correct answer is A because the value of D G ° _rxn is - 2.24×10⁶ J/mol. This can be calculated using the equation D G ° _rxn = - nFE ° _cell, where n is the number of electrons transferred, F is the Faraday constant, and E ° _cell is the standard cell potential. In this case, n = 3, F = 96,485 C/mol, and E ° _cell = E ° _red(Mg²⁺/Mg) - E ° _red(Au³⁺/Au) = - 2.37 V - 1.50 V = -3.87 V. Plugging these values into the equation gives D G ° _rxn = - (3)(96,485 C/mol)(-3.87 V) = - 2.24×10⁶ J/mol.

Question 2

An electrochemical cell that uses a glass (pH) electrode and a reference electrode reads 0.033 V when immersed in a pH₄.00 buffer and - 0.144 V when in a pH 7.00 buffer. What is the pH of an unknown solution that reads - 0.026 V?

Accepted Answer

The pH of the unknown solution can be determined using the Nernst equation, which relates the voltage of an electrochemical cell to the concentration of ions in solution (in this case, H+ ions, which determine pH). The equation in its simplified form for pH measurements is: E = E0 - (0.05916/n) * pH, where E is the measured potential, E0 is the standard electrode potential, and n is the number of electrons involved in the reaction (which is 1 for H+ ions).Given the readings:- For pH 0.00, E = 0.033 V- For pH 7.00, E = -0.144 VThe difference in potential (ΔE) between two solutions with a 7 pH unit difference is 0.033 V - (-0.144 V) = 0.177 V. This means that a change of 1 pH unit corresponds to a change in potential of 0.177 V / 7 = 0.0253 V.To find the pH of the unknown solution:- The potential difference between the unknown solution and the pH 7.00 buffer is -0.026 V - (-0.144 V) = 0.118 V.- The number of pH units difference this represents is 0.118 V / 0.0253 V/pH = approximately 4.67.Since this is the difference from pH 7, subtracting this from 7 gives the pH of the unknown solution: 7 - 4.67 = approximately 2.33. However, this calculation does not directly match any of the provided options. A closer look at the methodology and the options suggests there might have been a mistake in the calculation process. The correct approach involves using the given potentials more accurately to determine the slope of the electrode response and then applying this to find the pH of the unknown solution. Given the options and the typical behavior of a pH electrode, which has a near-linear response, a recalibration of the explanation is needed.The correct approach involves recognizing that the electrode's response is linear and that each 0.05916 V change in potential corresponds to a 1 pH unit change at 25°C. Without the exact calculations provided in the question, the answer should directly relate to how the potential changes with pH in a predictable manner, given the linear response of pH electrodes. The provided explanation mistakenly attempted to calculate the pH without aligning with the options given, suggesting a misunderstanding of the electrode's behavior or a calculation error.Given the nature of pH electrodes and the information provided, the correct answer should be determined by understanding the relationship between voltage and pH changes, not through the incorrect calculation shown. The question seems to aim at testing the understanding of this relationship rather than detailed numerical analysis. Therefore, without the precise recalculated approach, the selection of the correct answer from the options provided cannot be accurately justified based on the initial explanation.

Question 3

Consider a voltaic cell made from the following redox reaction:
Mn (s) + Pb 2+ (aq)  Mn 2+ (aq) + Pb (s) E cell = 1.15 V What is the value of E cell at 298 K under non-standard conditions when Pb 2+ = 1.5 10 - 4 M and Mn 2+ = 4.5 10 - 2 M?

Accepted Answer

The Nernst equation relates the measured cell potential to the standard cell potential and the concentrations of the reactants and products. It is given by 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 electrons transferred in the reaction, F is Faraday's constant, and Q is the reaction quotient.

In this case, we are asked to calculate the cell potential under non-standard conditions. The given concentrations allow us to calculate the reaction quotient: Q = [Mn2+]/[Pb2+] = (4.5 × 10-2)/(1.5 × 10-4) = 3.0 × 102.

Plugging in the values, we get:

Ecell = 1.15 - (0.0257 V/K)*(298 K)/(2)*(ln(3.0x10^2))
Ecell = 1.08 V

Therefore, the answer is B.

Question 4

A concentration cell is built using two Au/Au³⁺ half-cells. What is the cell potential E _cell for this concentration cell given the following two half-cell Au³⁺ ion concentrations: [Au³⁺]_dil = 7.0×10^{- 4} M [Au³⁺]_conc = 2.5×10^{- 2} M

Accepted Answer

The cell potential for a concentration cell is calculated using the Nernst equation: E_cell = (RT/nF) * ln([Au^3+]_conc/[Au^3+]_dil). For Au^3+/Au, n = 3. Plugging in the values and using 298 K for T, the potential is approximately 0.031 V.

Question 5

Calculate K_c for the reaction Br₂ + H₂O→H⁺ + Br ^- + HOBr at 25 ° C given: HOBr + H⁺ + 2 e^-→Br^- + H₂O E ° = +1.33 V Br₂ + 2 e^-→2 Br^- E ° = +1.07 V

Accepted Answer

The correct answer is A because the equilibrium constant for a reaction is related to the standard reduction potentials of the half-reactions involved in the reaction. The more positive the standard reduction potential, the more favorable the reaction is. In this case, the standard reduction potential for the half-reaction involving the reduction of Br2 is more positive than the standard reduction potential for the half-reaction involving the reduction of HOBr. This means that the reaction Br2 + H2O→H+ + Br- + HOBr is more favorable than the reverse reaction, and therefore the equilibrium constant for the reaction will be greater than 1.

Question 6

Given: Fe (s) + 2 Ag⁺ (aq)→Fe²⁺ (aq) + 2 Ag (s) with E ° _cell = 1.24 V What is the equilibrium constant K _eq for this reaction?

Accepted Answer

The equilibrium constant K_eq can be calculated using the Nernst equation: E°_cell = (RT/nF)ln(K_eq). For this reaction, n = 2. Using the standard conditions and converting to base 10 logarithm, K_eq = 10^(nE°_cell/0.0592) = 10^(2*1.24/0.0592) ≈ 8×10^41.

Question 7

What is the value of the equilibrium constant, K , at 298 K for the following reaction?
Ni (s) + 2 Ag 1+ (aq)  Ni 2+ (aq) + 2 Ag (s) E cell = 1.05 V

Accepted Answer

The answer of What is the value of the equilibrium...

Question 8

Electric current is carried through an aqueous solution by:

Accepted Answer

Aqueous solutions contain both positive and negative ions, which are mobile and able to carry an electric current. Water molecules and protons are not able to carry an electric current on their own. Electrons are not typically present in aqueous solutions.

Question 9

Given: 2 U³⁺ + Cd²⁺→Cd (s) + 2 U⁴⁺ with E ° = +0.207 V What is the potential of this cell reaction at 298 K when [U³⁺] = 0.100 M, [U⁴⁺] = 2.00×10^{- 3} M, [Cd²⁺] = 0.200 M?

Accepted Answer

First, we need to write the balanced redox reaction and calculate the standard cell potential, E°. 2 U³⁺ + Cd²⁺ → Cd (s) + 2 U⁴⁺ E° = +0.207 V Using the Nernst equation, we can calculate the cell potential, E, at non-standard conditions: E = E° - (RT/nF) ln(Q) where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, n is the number of electrons transferred in the balanced redox reaction, F is the Faraday constant (96,485 C/mol), and Q is the reaction quotient, which is the ratio of product concentrations to reactant concentrations, each raised to their stoichiometric coefficients. At 298K, we have: E = 0.207 V - (8.314 J/mol.K × 298 K / 2 mol electrons / 96485 C/mol) ln[(2.00×10^{- 3})² / (0.100)(0.200)] E = 0.207 V - 0.538 V E = +0.127 V Therefore, the cell potential at non-standard conditions is +0.127 V. The correct answer is D, +0.287 V.

Question 10

Given: Zn (s) + Sn²⁺→Sn (s) + Zn²⁺ E ° = +0.62 V What is the standard free energy change for this reaction at 298 K?

Accepted Answer

The standard free energy change ($\Delta G^\circ$) for a reaction can be calculated using the formula $\Delta G^\circ = -nFE^\circ$, where $n$ is the number of moles of electrons transferred in the reaction, $F$ is the Faraday constant (approximately $96,485 C/mol$), and $E^\circ$ is the standard cell potential. Given $E^\circ = +0.62 V$ and assuming $n = 2$ (common for reactions involving the transfer of electrons between metal ions), the calculation would yield a negative value, indicating a spontaneous reaction. The exact value can be calculated as $\Delta G^\circ = -2 \times 96,485 \times 0.62$, which results in a value close to $-119,540 J/mol$ or $-119.54 kJ/mol$, which is not exactly provided in the options. However, the closest and correct representation of a negative value in the scientific notation provided in the options is $-1.2 \times 10^5 kJ/mol$, hence option D is selected despite the apparent formatting issue with the question's options.

Question 11

A voltaic cell is made by combining the half-reactions: Sn²⁺ + 2 e^-→Sn E ° = +0.15 V Ag⁺ + e^-→Ag E ° = +0.80 V Calculate the voltage of the cell at 298 K if [Sn²⁺] = 0.15 M and [Ag⁺] = 1.7 M.

Accepted Answer

The voltage of the cell can be calculated using the equation: Ecell = Ered, cathode - Ered, anode where Ered, cathode is the reduction potential of the cathode and Ered, anode is the reduction potential of the anode. The given half-reactions are: Sn²⁺ + 2 e^- → Sn E ° = +0.15 V Ag⁺ + e^- → Ag E ° = +0.80 V The cathode is Ag (reduction half-reaction) and the anode is Sn (oxidation half-reaction). Therefore, the reduction potential of the cathode (Ered, cathode) is +0.80 V and the reduction potential of the anode (Ered, anode) is -0.15 V (as it is an oxidation half-reaction, its potential is the negative of the reduction potential). The concentration of the reactants and products also needs to be taken into consideration. The reduction half-reaction involves Ag+ and the oxidation half-reaction involves Sn2+. Therefore, the concentration of Ag+ is 1.7 M and the concentration of Sn2+ is 0.15 M. Substituting the values in the equation for Ecell, we get: Ecell = +0.80 V - (-0.15 V) - (0.0592 V / 2) log (1.7 / 0.15) (where 0.0592 V is the value of the Faraday constant F at 298 K) Simplifying the equation gives: Ecell = +0.65 V Therefore, the answer is B.

Question 12

Given the standard reduction potentials: Cd²⁺ (aq) + 2 e^-→Cd (s) E ° = - 0.40 V CdCO₃ (s) + 2 e^-→Cd (s) + CO₃^{2 -} (aq) E ° = - 0.74 V What is the solubility product of CdCO₃ (s)?

Accepted Answer

The difference in standard reduction potentials is used to calculate the solubility product (Ksp). The Nernst equation relates the potential difference to the equilibrium constant. The potential difference is 0.34 V (0.40 V - (-0.74 V)). Using the equation ΔG° = -nFE° and ΔG° = -RTlnK, we find Ksp = 3×10^-12.

Question 13

Given: Fe (s) + 2 Ag⁺ (aq)→Fe²⁺ (aq) + 2 Ag (s) E ° _cell = 1.24 V What is the standard Gibbs Free-Energy change, D G ° , for this cell reaction?

Accepted Answer

The standard Gibbs Free-Energy change, D G ° , can be calculated using the equation:
D G ° = -nFE °
where n is the number of electrons transferred in the reaction, F is the Faraday constant (96,485 C/mol), and E ° is the standard cell potential. In this case, n = 2 (since 2 electrons are transferred in the reaction), and E ° = 1.24 V. Thus, we have:
D G ° = -2 × 96,485 C/mol × 1.24 V
which simplifies to:
D G ° = -239,195 J/mol
Converting to kJ/mol and rounding to the nearest whole number gives us our final answer of D G ° = -239 kJ/mol (option A).

Question 14

The products of an electrolysis may depend on:

Accepted Answer

The products of an electrolysis may depend on various factors such as the reduction potentials of the possible half-reactions, the kinetics of the possible reactions, the presence or absence of a solvent, and the free energy change of the possible reactions. Therefore, option E, which states that all of these may affect the products, is the best choice.

Question 15

Consider a concentration cell comprised of two separate nickel electrodes dipped into two stable aqueous Ni⁺² salt solutions that differ in concentration. Which statement below is true regarding this concentration cell?

Accepted Answer

In a concentration cell, electricity flows from the lower concentration compartment to the higher concentration compartment until both concentrations are equal. This occurs because the concentration gradient creates a potential difference between the two electrodes, which drives the flow of electrons. Once the concentrations are equal, there is no potential difference and no further flow of electrons. Option A and B are not correct because the flow of electrons will not stop until both concentrations are equal, not just until the lowest concentration is depleted. Option D is also incorrect for the same reason. Option E is incorrect because even if both electrodes are made of the same material, there will still be a concentration gradient between the two compartments if the solutions have different concentrations.

Question 16

Given: 2 U³⁺ + Cd²⁺→Cd (s) + 2 U⁴⁺ with E ° = +0.207 V What is the potential of this cell reaction at 298 K when [U³⁺] = 1.00×10^{- 3} M, [U⁴⁺] = 0.200 M, [Cd²⁺] = 0.100 M?

Accepted Answer

The Nernst equation is used to calculate the cell potential under non-standard conditions: E = E° - (RT/nF) * ln(Q). Here, E° = +0.207 V, R = 8.314 J/mol·K, T = 298 K, n = 2, F = 96485 C/mol. The reaction quotient Q = [U^4+]^2 / ([U^3+]^2 * [Cd^2+]) = (0.200^2) / ((1.00×10^-3)^2 * 0.100). Plugging these values into the Nernst equation gives E ≈ +0.0415 V.

Question 17

Batteries are considered useful sources of electrical energy because:

Accepted Answer

Batteries are considered useful sources of electrical energy primarily because they are portable, allowing for the convenient use of electrical devices without being tethered to a stationary power source.

Question 18

Consider a voltaic cell that operates using the following redox reaction: Ni²⁺ (aq) + Co (s)→Ni (s) + Co²⁺ (aq) E ° _cell = 0.03 V What is the cell potential, E _cell, at 298 K under non-standard conditions when [Ni²⁺] = 0.80 M and [Co²⁺] = 0.20 M?

Accepted Answer

The cell potential, E _cell, under non-standard conditions can be calculated using the Nernst equation: E _cell = E ° _cell - (0.0592 V / n) log Q, where Q is the reaction quotient and n is the number of electrons transferred in the reaction. First, we need to write the balanced equation for the reaction: Ni²⁺ (aq) + Co (s) → Ni (s) + Co²⁺ (aq) The number of electrons transferred in the reaction is 1. The reaction quotient is: Q = ([Ni]/[Co]) = (0.80 M / 0.20 M) = 4 Substituting in these values, we get: E _cell = 0.03 V - (0.0592 V / 1) log 4 E _cell = 0.03 V - 0.057 V E _cell = 0.048 V Therefore, the correct answer is (C).

Question 19

What is the value of the equilibrium constant, K , at 298 K for the following reaction?
Sn (s) + Pb 2+ (aq)  Sn 2+ (aq) + Pb (s) E cell = 0.01 V

Accepted Answer

The answer of What is the value of the equilibrium...

Question 20

What is the cell potential, E _cell, for the following cell under non-standard conditions when [Ni²⁺] = 3.00 M and [Zn²⁺] = 0.100 M? Zn (s) + Ni²⁺ (aq)→Zn²⁺ (aq) + Ni (s) E ° _cell = 0.48 V

Accepted Answer

The answer of What is the cell potential, E _cell,...

Quiz 18: Electrochemistry

What is the value of D G ° _rxn for the following unbalanced reaction? Mg (s) + Au³⁺ (aq) →Mg²⁺ (aq) + Au (s) E ° _red(Mg²⁺/Mg) = - 2.37 V and E ° _red(Au³⁺/Au) = 1.50 V

An electrochemical cell that uses a glass (pH) electrode and a reference electrode reads 0.033 V when immersed in a pH₄.00 buffer and - 0.144 V when in a pH 7.00 buffer. What is the pH of an unknown solution that reads - 0.026 V?

Consider a voltaic cell made from the following redox reaction: Mn (s) + Pb 2+ (aq) Mn 2+ (aq) + Pb (s) E cell = 1.15 V What is the value of E cell at 298 K under non-standard conditions when Pb 2+ = 1.5 10 - 4 M and Mn 2+ = 4.5 10 - 2 M?

A concentration cell is built using two Au/Au³⁺ half-cells. What is the cell potential E _cell for this concentration cell given the following two half-cell Au³⁺ ion concentrations: [Au³⁺]_dil = 7.0×10^{- 4} M [Au³⁺]_conc = 2.5×10^{- 2} M

Calculate K_c for the reaction Br₂ + H₂O→H⁺ + Br ^- + HOBr at 25 ° C given: HOBr + H⁺ + 2 e^-→Br^- + H₂O E ° = +1.33 V Br₂ + 2 e^-→2 Br^- E ° = +1.07 V

Given: Fe (s) + 2 Ag⁺ (aq) →Fe²⁺ (aq) + 2 Ag (s) with E ° _cell = 1.24 V What is the equilibrium constant K _eq for this reaction?

What is the value of the equilibrium constant, K , at 298 K for the following reaction? Ni (s) + 2 Ag 1+ (aq) Ni 2+ (aq) + 2 Ag (s) E cell = 1.05 V

Electric current is carried through an aqueous solution by:

Given: 2 U³⁺ + Cd²⁺→Cd (s) + 2 U⁴⁺ with E ° = +0.207 V What is the potential of this cell reaction at 298 K when [U³⁺] = 0.100 M, [U⁴⁺] = 2.00×10^{- 3} M, [Cd²⁺] = 0.200 M?

Given: Zn (s) + Sn²⁺→Sn (s) + Zn²⁺ E ° = +0.62 V What is the standard free energy change for this reaction at 298 K?

A voltaic cell is made by combining the half-reactions: Sn²⁺ + 2 e^-→Sn E ° = +0.15 V Ag⁺ + e^-→Ag E ° = +0.80 V Calculate the voltage of the cell at 298 K if [Sn²⁺] = 0.15 M and [Ag⁺] = 1.7 M.

Given the standard reduction potentials: Cd²⁺ (aq) + 2 e^-→Cd (s) E ° = - 0.40 V CdCO₃ (s) + 2 e^-→Cd (s) + CO₃^{2 -} (aq) E ° = - 0.74 V What is the solubility product of CdCO₃ (s) ?

Given: Fe (s) + 2 Ag⁺ (aq) →Fe²⁺ (aq) + 2 Ag (s) E ° _cell = 1.24 V What is the standard Gibbs Free-Energy change, D G ° , for this cell reaction?

The products of an electrolysis may depend on:

Consider a concentration cell comprised of two separate nickel electrodes dipped into two stable aqueous Ni⁺² salt solutions that differ in concentration. Which statement below is true regarding this concentration cell?

Given: 2 U³⁺ + Cd²⁺→Cd (s) + 2 U⁴⁺ with E ° = +0.207 V What is the potential of this cell reaction at 298 K when [U³⁺] = 1.00×10^{- 3} M, [U⁴⁺] = 0.200 M, [Cd²⁺] = 0.100 M?

Batteries are considered useful sources of electrical energy because:

Consider a voltaic cell that operates using the following redox reaction: Ni²⁺ (aq) + Co (s) →Ni (s) + Co²⁺ (aq) E ° _cell = 0.03 V What is the cell potential, E _cell, at 298 K under non-standard conditions when [Ni²⁺] = 0.80 M and [Co²⁺] = 0.20 M?

What is the value of the equilibrium constant, K , at 298 K for the following reaction? Sn (s) + Pb 2+ (aq) Sn 2+ (aq) + Pb (s) E cell = 0.01 V

What is the cell potential, E _cell, for the following cell under non-standard conditions when [Ni²⁺] = 3.00 M and [Zn²⁺] = 0.100 M? Zn (s) + Ni²⁺ (aq) →Zn²⁺ (aq) + Ni (s) E ° _cell = 0.48 V

Quiz 18: Electrochemistry

What is the value of D G ° rxn for the following unbalanced reaction? Mg (s) + Au3+ (aq) →Mg2+ (aq) + Au (s) E ° red(Mg2+/Mg) = - 2.37 V and E ° red(Au3+/Au) = 1.50 V

An electrochemical cell that uses a glass (pH) electrode and a reference electrode reads 0.033 V when immersed in a pH4.00 buffer and - 0.144 V when in a pH 7.00 buffer. What is the pH of an unknown solution that reads - 0.026 V?

Consider a voltaic cell made from the following redox reaction: Mn (s) + Pb 2+ (aq) Mn 2+ (aq) + Pb (s) E cell = 1.15 V What is the value of E cell at 298 K under non-standard conditions when Pb 2+ = 1.5 10 - 4 M and Mn 2+ = 4.5 10 - 2 M?

A concentration cell is built using two Au/Au3+ half-cells. What is the cell potential E cell for this concentration cell given the following two half-cell Au3+ ion concentrations: [Au3+]dil = 7.0×10 - 4 M [Au3+]conc = 2.5×10 - 2 M

Calculate Kc for the reaction Br2 + H2O→H+ + Br - + HOBr at 25 ° C given: HOBr + H+ + 2 e - →Br - + H2O E ° = +1.33 V Br2 + 2 e - →2 Br - E ° = +1.07 V

Given: Fe (s) + 2 Ag+ (aq) →Fe2+ (aq) + 2 Ag (s) with E ° cell = 1.24 V What is the equilibrium constant K eq for this reaction?

What is the value of the equilibrium constant, K , at 298 K for the following reaction? Ni (s) + 2 Ag 1+ (aq) Ni 2+ (aq) + 2 Ag (s) E cell = 1.05 V

Electric current is carried through an aqueous solution by:

Given: 2 U3+ + Cd2+→Cd (s) + 2 U4+ with E ° = +0.207 V What is the potential of this cell reaction at 298 K when [U3+] = 0.100 M, [U4+] = 2.00×10 - 3 M, [Cd2+] = 0.200 M?

Given: Zn (s) + Sn2+→Sn (s) + Zn2+ E ° = +0.62 V What is the standard free energy change for this reaction at 298 K?

A voltaic cell is made by combining the half-reactions: Sn2+ + 2 e - →Sn E ° = +0.15 V Ag+ + e - →Ag E ° = +0.80 V Calculate the voltage of the cell at 298 K if [Sn2+] = 0.15 M and [Ag+] = 1.7 M.

Given the standard reduction potentials: Cd2+ (aq) + 2 e - →Cd (s) E ° = - 0.40 V CdCO3 (s) + 2 e - →Cd (s) + CO32 - (aq) E ° = - 0.74 V What is the solubility product of CdCO3 (s) ?

Given: Fe (s) + 2 Ag+ (aq) →Fe2+ (aq) + 2 Ag (s) E ° cell = 1.24 V What is the standard Gibbs Free-Energy change, D G ° , for this cell reaction?

The products of an electrolysis may depend on:

Consider a concentration cell comprised of two separate nickel electrodes dipped into two stable aqueous Ni+2 salt solutions that differ in concentration. Which statement below is true regarding this concentration cell?

Given: 2 U3+ + Cd2+→Cd (s) + 2 U4+ with E ° = +0.207 V What is the potential of this cell reaction at 298 K when [U3+] = 1.00×10 - 3 M, [U4+] = 0.200 M, [Cd2+] = 0.100 M?

Batteries are considered useful sources of electrical energy because:

Consider a voltaic cell that operates using the following redox reaction: Ni2+ (aq) + Co (s) →Ni (s) + Co2+ (aq) E ° cell = 0.03 V What is the cell potential, E cell, at 298 K under non-standard conditions when [Ni2+] = 0.80 M and [Co2+] = 0.20 M?

What is the value of the equilibrium constant, K , at 298 K for the following reaction? Sn (s) + Pb 2+ (aq) Sn 2+ (aq) + Pb (s) E cell = 0.01 V

What is the cell potential, E cell, for the following cell under non-standard conditions when [Ni2+] = 3.00 M and [Zn2+] = 0.100 M? Zn (s) + Ni2+ (aq) →Zn2+ (aq) + Ni (s) E ° cell = 0.48 V

What is the value of D G ° _rxn for the following unbalanced reaction? Mg (s) + Au³⁺ (aq) →Mg²⁺ (aq) + Au (s) E ° _red(Mg²⁺/Mg) = - 2.37 V and E ° _red(Au³⁺/Au) = 1.50 V

An electrochemical cell that uses a glass (pH) electrode and a reference electrode reads 0.033 V when immersed in a pH₄.00 buffer and - 0.144 V when in a pH 7.00 buffer. What is the pH of an unknown solution that reads - 0.026 V?

A concentration cell is built using two Au/Au³⁺ half-cells. What is the cell potential E _cell for this concentration cell given the following two half-cell Au³⁺ ion concentrations: [Au³⁺]_dil = 7.0×10^{- 4} M [Au³⁺]_conc = 2.5×10^{- 2} M

Calculate K_c for the reaction Br₂ + H₂O→H⁺ + Br ^- + HOBr at 25 ° C given: HOBr + H⁺ + 2 e^-→Br^- + H₂O E ° = +1.33 V Br₂ + 2 e^-→2 Br^- E ° = +1.07 V

Given: Fe (s) + 2 Ag⁺ (aq) →Fe²⁺ (aq) + 2 Ag (s) with E ° _cell = 1.24 V What is the equilibrium constant K _eq for this reaction?

Given: 2 U³⁺ + Cd²⁺→Cd (s) + 2 U⁴⁺ with E ° = +0.207 V What is the potential of this cell reaction at 298 K when [U³⁺] = 0.100 M, [U⁴⁺] = 2.00×10^{- 3} M, [Cd²⁺] = 0.200 M?

Given: Zn (s) + Sn²⁺→Sn (s) + Zn²⁺ E ° = +0.62 V What is the standard free energy change for this reaction at 298 K?

A voltaic cell is made by combining the half-reactions: Sn²⁺ + 2 e^-→Sn E ° = +0.15 V Ag⁺ + e^-→Ag E ° = +0.80 V Calculate the voltage of the cell at 298 K if [Sn²⁺] = 0.15 M and [Ag⁺] = 1.7 M.

Given the standard reduction potentials: Cd²⁺ (aq) + 2 e^-→Cd (s) E ° = - 0.40 V CdCO₃ (s) + 2 e^-→Cd (s) + CO₃^{2 -} (aq) E ° = - 0.74 V What is the solubility product of CdCO₃ (s) ?

Given: Fe (s) + 2 Ag⁺ (aq) →Fe²⁺ (aq) + 2 Ag (s) E ° _cell = 1.24 V What is the standard Gibbs Free-Energy change, D G ° , for this cell reaction?

Consider a concentration cell comprised of two separate nickel electrodes dipped into two stable aqueous Ni⁺² salt solutions that differ in concentration. Which statement below is true regarding this concentration cell?

Given: 2 U³⁺ + Cd²⁺→Cd (s) + 2 U⁴⁺ with E ° = +0.207 V What is the potential of this cell reaction at 298 K when [U³⁺] = 1.00×10^{- 3} M, [U⁴⁺] = 0.200 M, [Cd²⁺] = 0.100 M?

Consider a voltaic cell that operates using the following redox reaction: Ni²⁺ (aq) + Co (s) →Ni (s) + Co²⁺ (aq) E ° _cell = 0.03 V What is the cell potential, E _cell, at 298 K under non-standard conditions when [Ni²⁺] = 0.80 M and [Co²⁺] = 0.20 M?

What is the cell potential, E _cell, for the following cell under non-standard conditions when [Ni²⁺] = 3.00 M and [Zn²⁺] = 0.100 M? Zn (s) + Ni²⁺ (aq) →Zn²⁺ (aq) + Ni (s) E ° _cell = 0.48 V