Gold exists in two common positive oxidation states, +1 and +3. The standard reduction potentials for these oxidation states are Au+1aq2 + e- ¡ Au1s2 Ered ° = +1.69 V Au3+1aq2 + 3 e- ¡ Au1s2 Ered ° = +1.50 V (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction 4 Au1s2 + 8 NaCN1aq2 + 2 H2O1l2 + O21g2 ¡ 4 Na3Au1CN2241aq2 + 4 NaOH1aq2 What is being oxidized, and what is being reduced in this reaction?
Ch.20 - Electrochemistry
Chapter 20, Problem 102a
A voltaic cell is constructed that uses the following half-cell reactions:
Cu+(aq) + e- → Cu(s)
I2(s) + 2 e- → 2 I-(aq)
The cell is operated at 298 K with [Cu+] = 0.25 M and [I-] = 0.035 M.
(a) Determine E for the cell at these concentrations.

1
Identify the oxidation and reduction half-reactions. In this case, Cu+ is reduced to Cu(s) and I2(s) is oxidized to I-.
Write the balanced overall cell reaction by combining the half-reactions. Ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction.
Use the Nernst equation to calculate the cell potential (E) at non-standard conditions. The Nernst equation is: E = E^0 - (RT/nF) * ln(Q), where E^0 is the standard cell potential, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is the Faraday constant (96485 C/mol), and Q is the reaction quotient.
Calculate the reaction quotient, Q, from the given concentrations. For the reaction Cu+ + I2 -> Cu + 2 I-, Q is calculated as Q = [I-]^2 / [Cu+].
Substitute the values of E^0, R, T, n, F, and Q into the Nernst equation to find the cell potential E.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Electrochemical Cells
Electrochemical cells, such as voltaic cells, convert chemical energy into electrical energy through redox reactions. In a voltaic cell, oxidation occurs at the anode and reduction at the cathode, allowing for the flow of electrons through an external circuit. Understanding the setup and function of these cells is crucial for analyzing their behavior under different conditions.
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Nernst Equation
The Nernst equation relates the cell potential (E) to the concentrations of the reactants and products in a redox reaction. It is expressed as 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 moles of electrons transferred, F is Faraday's constant, and Q is the reaction quotient. This equation is essential for calculating the cell potential under non-standard conditions.
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Reaction Quotient (Q)
The reaction quotient (Q) is a measure of the relative concentrations of reactants and products at any point in a reaction. It is calculated using the formula Q = [products]/[reactants], with each concentration raised to the power of its stoichiometric coefficient. In the context of the Nernst equation, Q helps determine how the cell potential changes as the reaction progresses and concentrations vary.
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Related Practice
Textbook Question
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Open Question
A voltaic cell is constructed from an Ni2+(aq) / Ni(s) half-cell and an Ag+(aq) / Ag(s) half-cell. The initial concentration of Ni2+(aq) in the Ni2+ - Ni half-cell is [Ni2+] = 0.0100 M. The initial cell voltage is +1.12 V. (a) By using data in Appendix E, calculate the standard emf of this voltaic cell.
Open Question
Will the concentration of Ni2+ in the Ni2+ - Ni half-cell increase or decrease as the cell operates?
Textbook Question
A voltaic cell is constructed that uses the following half-cell reactions:
Cu+(aq) + e- → Cu(s)
I2(s) + 2 e- → 2 I-(aq)
The cell is operated at 298 K with [Cu+] = 0.25 M and [I-] = 0.035 M.
(b) Which electrode is the anode of the cell?
(c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions?
Open Question
Using data from Appendix E, calculate the equilibrium constant for the disproportionation of the copper(I) ion at room temperature: 2 Cu+(aq) ⇌ Cu2+(aq) + Cu(s).
Textbook Question
(b) Given
the following reduction potentials, calculate the standard
emf of the cell:
Cd1OH221s2 + 2 e- ¡ Cd1s2 + 2 OH-1aq2
E°red = -0.76 V
NiO1OH21s2 + H2O1l2 + e- ¡ Ni1OH221s2 + OH-1aq2
E°red = +0.49 V
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