Problem 29 - Chemistry and Chemical Reactivity

Question

One half-cell in a voltaic cell is constructed from a silver wire electrode in a AgNO \(_{3}\) solution of unknown concentration. The other half-cell consists of a zinc electrode in a \(1.0 \mathrm{M}\) solution of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} .\) A potential of \(1.48 \mathrm{V}\) is measured for this cell. Use this information to calculate the concentration of \(\mathrm{Ag}^{+}(\mathrm{aq})\)

Step-by-Step Solution

Answer

The concentration of \( \mathrm{Ag}^+ \) is approximately 22.5 M.

Step 1: Identify the standard electrode potentials

Identify the standard electrode potentials for both silver (Ag) and zinc (Zn). From standard tables, we know:\(E^{º}_{Ag^+/Ag} = +0.80 \text{ V}\)\(E^{º}_{Zn^{2+}/Zn} = -0.76 \text{ V}\)

Step 2: Determine the cell potential equation

For the cell, the overall cell potential (Eº) is the difference between the cathode and anode potentials. The silver electrode is the cathode, and the zinc electrode is the anode. The equation is: \(E^{º}_{cell} = E^{º}_{Ag^+/Ag} - E^{º}_{Zn^{2+}/Zn} = 0.80 - (-0.76) = 1.56 \text{ V}\)

Step 3: Use the Nernst Equation

Given the measured cell potential of 1.48 V and the standard cell potential of 1.56 V, we use the Nernst equation to find the concentration of \(\mathrm{Ag}^+\). The Nernst equation is:\[E_{cell} = E^{º}_{cell} - \frac{RT}{nF} \ln Q\] Where \(R\) is the gas constant (8.314 J/mol·K), \(T\) is the temperature in Kelvin (298 K for standard conditions), \(n\) is the number of moles of electrons transferred in the reaction (2 for this reaction), \(F\) is Faraday's constant (96485 C/mol), and \(Q\) is the reaction quotient \(\frac{1}{[Ag^+]^2}\) since: \(Zn + 2Ag^+ \rightleftharpoons Zn^{2+} + 2Ag\)

Step 4: Rearrange and solve the Nernst Equation for [Ag+]

Plug in the values into the Nernst equation and solve for \([Ag^+]\):\[1.48 = 1.56 - \frac{0.0257}{2} \ln \left(\frac{1}{[Ag^+]^2}\right)\] Solving for \([Ag^+]\): \(0.08 = -0.01285 \ln \left(\frac{1}{[Ag^+]^2}\right)\) Divide both sides by \(-0.01285\):\(-6.23 = \ln \left(\frac{1}{[Ag^+]^2}\right)\) Taking the exponential of both sides to solve for \([Ag^+]^2\):\(\frac{1}{[Ag^+]^2} = e^{-6.23}\) \([Ag^+] = \sqrt{\frac{1}{e^{-6.23}}}\)Calculate this to find \([Ag^+]\).

Step 5: Calculate [Ag+]

Compute the value found in the previous step.\[\ln \left(\frac{1}{[Ag^+]^2}\right) = -6.23 \rightarrow [Ag^+]^2 = e^{6.23} \approx 508\] \([Ag^+] = \sqrt{508} \approx 22.5\, \text{M}\)\, so \( \text{concentration of Ag}^+\) is approximately 22.5 M.

Key Concepts

Nernst Equation

In the world of electrochemistry, the Nernst Equation is an invaluable tool for understanding how electrode potentials change under non-standard conditions. At its core, this equation allows us to relate the potential of an electrochemical cell at any given concentration to its standard cell potential. Its general form is: \[ E_{cell} = E^{º}_{cell} - \frac{RT}{nF} \ln Q \]Where:

• **\(E_{cell}\):** The cell potential at non-standard conditions.
• **\(E^{º}_{cell}\):** The standard cell potential.
• **\(R\):** The universal gas constant (8.314 J/mol·K).
• **\(T\):** The temperature in Kelvin.
• **\(n\):** The number of moles of electrons transferred in the balanced equation.
• **\(F\):** Faraday's constant (96485 C/mol).
• **\(Q\):** The reaction quotient, reflecting the ratio of product and reactant concentrations.

By rearranging and solving this equation, we can deduce unknown concentrations or cell potentials. In our specific exercise example, we used it to determine the concentration of \( \mathrm{Ag}^+ \) ions by finding the relation between measured and standard potential differences.

Electrode Potentials

Electrode potentials are central to understanding how voltaic cells operate. They arise from the tendency of a half-reaction to occur as either a reduction or oxidation at an electrode. Numbers, drawn from standard conditions, are typically expressed in volts and can be found in electrochemical tables.In standard configurations, each metal is compared against the standard hydrogen electrode (SHE). The potential for the reduction of an ion to its element (like \( \mathrm{Ag}^+ \to \mathrm{Ag} \)) is given a specific value:

• The standard electrode potential for silver is \( E^{º}_{Ag^+/Ag} = +0.80 \text{ V} \).
• For zinc, the potential is \( E^{º}_{Zn^{2+}/Zn} = -0.76 \text{ V} \).

These values allow us to calculate the "driving force" of an entire cell, determined by the difference between the cathode and the anode potentials. This inspires the flow of ions from a higher to a lower potential energy state, driving the voltaic cell's reaction.

Voltaic Cells

Voltaic cells, or galvanic cells, convert chemical energy into electrical energy through spontaneous redox reactions. Composed of two half-cells, each contains an electrode and an electrolyte. Here's a simplified breakdown of their workings:

• Each half-cell produces a specific potential, based on the electrode material and the solution.
• The anode is the electrode where oxidation occurs, releasing electrons.
• The cathode is where reduction takes place, gaining electrons.
• An external circuit connects these electrodes, allowing for electron flow and performing electrical work.
• Anions and cations flow through a salt bridge to maintain charge balance.

For the provided exercise, our voltaic cell consists of a silver (Ag) and a zinc (Zn) electrode. With zinc ions in the solution as the anode producing electrons, and silver ions at the cathode collecting them, a definite voltage develops. This is the potential we measure to deduce the concentrations, as described in the exercise solution.