Question

Follow the directions in Question 17 for a salt bridge cell in which the anode is a platinum rod immersed in an aqueous solution of sodium iodide containing solid iodine crystals. The cathode is another platinum rod immersed in an aqueous solution of sodium bromide with bromine liquid.

Short answer

Answer: In this electrochemical cell, sodium iodide reacts with bromine to form iodine and sodium bromide, with an overall cell reaction of 2I⁻(aq) + Br2(l) → I2(s) + 2Br⁻(aq). The cell potential under standard conditions is 0.55 V.

Step-by-step solution

Identify the half-reactions

We need to recognize the oxidation and reduction half-reactions that take place at the anode and the cathode, respectively. At the anode, the iodide ion (I-) is oxidized, while at the cathode, the bromine molecule (Br2) is reduced. The half-reactions are: 1. Anode (oxidation): \(\textrm{I}^-_\textrm{(aq)} \rightarrow \frac{1}{2} \textrm{I}_\textrm{2(s)} + \textrm{e}^-\) 2. Cathode (reduction): \(\textrm{Br}_\textrm{2(l)} + 2\textrm{e}^- \rightarrow 2\textrm{Br}^-_\textrm{(aq)}\)

Calculate the standard electrode potentials

To determine the cell potential, we need to calculate the standard electrode potentials (\(E^\circ\)) for each half-reaction. The standard electrode potentials can be found in tables or a textbook, and they are: 1. Anode (oxidation): \(E^\circ_\textrm{I-/I2} = +0.54 \ \textrm{V}\) 2. Cathode (reduction): \(E^\circ_\textrm{Br2/Br-} = +1.09 \ \textrm{V}\)

Calculate the cell potential

The cell potential (\(E_{cell}\)) can be calculated using the formula: \(E_{cell} = E^\circ_\textrm{cathode} - E^\circ_\textrm{anode}\) Substituting the standard electrode potentials, we get: \(E_{cell} = 1.09 \ \textrm{V} - 0.54 \ \textrm{V} = 0.55\ \textrm{V}\)

Write the balanced equation for the overall cell reaction

To find the balanced equation for the overall cell reaction, we need to add the anode and the cathode half-reactions, ensuring that the number of electrons is the same for both reactions. Since the anode half-reaction releases one electron and the cathode half-reaction consumes two electrons, we must multiply the anode half-reaction by 2 to balance the electrons: 2 x \(\textrm{I}^-_\textrm{(aq)} \rightarrow \textrm{I}_\textrm{2(s)} + 2\textrm{e}^-\) Adding the balanced half-reactions and canceling out the electrons: \(2\textrm{I}^-_\textrm{(aq)} + \textrm{Br}_\textrm{2(l)} \rightarrow \textrm{I}_\textrm{2(s)} + 2\textrm{Br}^-_\textrm{(aq)}\) The overall cell reaction is: \(\textrm{2I}^-_\textrm{(aq)} + \textrm{Br}_\textrm{2(l)} \rightarrow \textrm{I}_\textrm{2(s)} + 2\textrm{Br}^-_\textrm{(aq)}\)

Interpret the results

The calculated cell potential (0.55 V) is positive, suggesting that the overall cell reaction is spontaneous under standard conditions. The balanced equation shows that the iodide ions are oxidized to iodine, and the bromine molecules are reduced to bromide ions. This indicates that sodium iodide reacts with bromine to form iodine and sodium bromide in this electrochemical cell.