Problem 76 - Chemistry and Chemical Reactivity

Question

A voltaic cell is constructed in which one half-cell consists of a silver wire in an aqueous solution of AgNO \(_{3} .\) The other half-cell consists of an inert platinum wire in an aqueous solution containing \(\mathrm{Fe}^{2+}(\mathrm{aq})\) and \(\mathrm{Fe}^{3+}(\mathrm{aq})\) (a) Calculate the voltage of the cell, assuming standard conditions. (b) Write the net ionic equation for the reaction occurring in the cell. (c) In this voltaic cell, which electrode is the anode and which is the cathode? (d) If \(\left[\mathrm{Ag}^{+}\right]\) is \(0.10 \mathrm{M},\) and \(\left[\mathrm{Fe}^{2+}\right]\) and \(\left[\mathrm{Fe}^{3+}\right]\) are both 1.0 M, what is the cell voltage? Is the net cell reaction still that used in part (a)? If not, what is the net reaction under the new conditions?

Step-by-Step Solution

Answer

(a) 0.03 V. (b) \( \mathrm{Ag}^+ + \mathrm{Fe}^{3+} \rightarrow \mathrm{Fe}^{2+} + \mathrm{Ag} \). (c) Anode: Fe, Cathode: Ag. (d) -0.0291 V; no, it is not spontaneous.

Step 1: Cell Reaction Equations

First, identify the half-reactions involved in the voltaic cell. The silver half-cell involves the reaction: \[ \mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag} \] The iron half-cell involves the reaction: \[ \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \] These form the overall cell reaction: \[ \mathrm{Ag}^+ + \mathrm{Fe}^{3+} \rightarrow \mathrm{Ag} + \mathrm{Fe}^{2+} \]

Step 2: Standard Cell Potential

Look up the standard reduction potentials for the half reactions. - Silver's standard reduction potential, \( E^0_{\mathrm{Ag}^+/\mathrm{Ag}} \), is +0.80 V. - Iron's standard reduction potential, \( E^0_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} \), is +0.77 V.Calculate the standard cell potential using:\[ E^0_{\text{cell}} = E^0_{ ext{cathode}} - E^0_{ ext{anode}} \]\[ E^0_{\text{cell}} = 0.80 \text{ V} - 0.77 \text{ V} = 0.03 \text{ V} \]

Step 3: Net Ionic Equation

The net ionic equation for the cell reaction is obtained by adding the two half-reactions:\[ \mathrm{Ag}^+ + \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Ag} + \mathrm{Fe}^{2+} + e^- \]The electrons cancel out, resulting in:\[ \mathrm{Ag}^+ + \mathrm{Fe}^{3+} \rightarrow \mathrm{Fe}^{2+} + \mathrm{Ag} \]

Step 4: Identifying Anode and Cathode

In a galvanic cell, the cathode is where reduction occurs, and the anode is where oxidation occurs. - For the Ag half-cell, \( \mathrm{Ag}^+ \) is reduced, making it the cathode.- For the Fe half-cell, \( \mathrm{Fe}^{2+} \) is oxidized (backwards reaction), making it the anode.

Step 5: Nernst Equation for Non-Standard Conditions

For non-standard conditions, use the Nernst equation:\[ E_{\text{cell}} = E^0_{\text{cell}} - \frac{RT}{nF} \ln Q \]Here, \( E^0_{\text{cell}} \) is 0.03 V from earlier, \( Q \) is the reaction quotient:\[ Q = \frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Ag}^+][\mathrm{Fe}^{3+}]} = \frac{1.0}{0.10 \times 1.0} = 10 \]Assuming room temperature (298 K), and \( n = 1 \):\[ E_{\text{cell}} = 0.03 - \frac{(8.314)(298)}{(1)(96485)} \ln(10) \]Calculate which gives:\[ E_{\text{cell}} = 0.03 - 0.0591 \times 1 = -0.0291 \text{ V} \]

Step 6: Checking Net Reaction Under New Conditions

With the calculated \( E_{\text{cell}} \) being negative under the given concentrations, the cell cannot operate as a voltaic cell anymore under these specific conditions. The net reaction remains the same, but it will not proceed spontaneously.

Key Concepts

Electrochemistry

Electrochemistry is all about the study of chemical reactions that involve electron transfer. In a voltaic cell, these reactions happen spontaneously, converting chemical energy into electrical energy.
This process occurs in two separate half-cells linked by a salt bridge, ensuring that ions can move freely while keeping the solutions apart. Understanding these reactions helps us harness energy efficiently, like in batteries.
In our given voltaic cell, we have a silver electrode and an iron electrode participating in these redox reactions. The key is to remember that the electron flow generates the electrical current we rely on.

Standard Cell Potential

The standard cell potential \(E^0_{\text{cell}}\) is a measure of the driving force behind an electrochemical reaction. It's calculated using standard reduction potentials, which are values given in reference tables.
For a reaction, one electrode acts as a cathode (where reduction occurs) and the other as an anode (where oxidation occurs).
The formula to calculate the standard cell potential is: \ E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} \.
In our case, the silver ion reduction has a potential of +0.80 V and iron ion reduction is +0.77 V. Subtracting gives us 0.03 V as the standard cell potential.

• Silver acts as the cathode because its potential is higher.
• Iron acts as the anode, with a lower potential.

Understanding this helps predict if a reaction will proceed using voltage.

Nernst Equation

The Nernst equation allows us to calculate cell potentials under non-standard conditions. Cell reactions don't always happen under perfect conditions like 1 M concentrations or 25°C temperature.
For more realistic environments, the equation modifies the potential based on the actual concentrations:
\[ E_{\text{cell}} = E^0_{\text{cell}} - \frac{RT}{nF} \ln Q \]
Where:

• R is the gas constant (8.314 J/mol·K),
• T is the temperature in Kelvin,
• n is the number of electrons exchanged,
• F is Faraday’s constant (96485 C/mol),
• Q is the reaction quotient, calculated from the ratio of product concentrations to reactant concentrations.

Applying these gives us the cell's real voltage, essential for understanding real-world reactions. Our exercise showed that for different concentrations, the voltage might even turn negative, stopping the reaction.

Net Ionic Equation

A net ionic equation represents only the ions and molecules that participate directly in a chemical reaction. In electrochemistry, it's important to describe the change happening in the cell.
For our voltaic cell, the separate half-reactions combine to form:\[ \mathrm{Ag}^+ + \mathrm{Fe}^{3+} \rightarrow \mathrm{Fe}^{2+} + \mathrm{Ag} \]
This equation shows only the species directly involved, ignoring spectator ions.

• Silver ion gains an electron to become silver metal.
• Iron(III) ion receives an electron, reducing to Iron(II).

Writing net ionic equations helps us focus on the essential parts of chemical reactions in the cell, clarifying the process.