Question

Show that for a combination of half-cell reactions that produce a standard reduction potential for a half-cell that is not directly observable, the standard reduction potential is $$E^{\circ}=\frac{\sum n_{i} E_{i}^{\circ}}{\sum n_{i}}$$ where \(n_{i}\) is the number of electrons in each half-reaction of potential \(E_{i}^{\circ} .\) Use the following half-reactions: $$ \begin{array}{c} \mathrm{H}_{5} \mathrm{IO}_{6}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{IO}_{3}^{-}(\mathrm{aq})+ \\ 3 \mathrm{H}_{2} \mathrm{O}(1) \quad E^{\circ}=1.60 \mathrm{V} \\ \mathrm{IO}_{3}^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq})+5 \mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{I}_{2}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(1) \\ E^{\circ}=1.19 \mathrm{V} \\ 2 \mathrm{HIO}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(1) \\ E^{\circ}=1.45 \mathrm{V} \\ \mathrm{I}_{2}(\mathrm{s})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(\mathrm{aq}) \quad \quad E^{\circ}=0.535 \mathrm{V} \end{array} $$ Calculate the standard reduction potential for $$ \mathrm{H}_{6} \mathrm{IO}_{6}+5 \mathrm{H}^{+}+2 \mathrm{I}^{-}+3 \mathrm{e}^{-} \longrightarrow $$ $$ \frac{1}{2} \mathrm{I}_{2}+4 \mathrm{H}_{2} \mathrm{O}=2 \mathrm{HIO} $$

Short answer

After performing the calculations, the standard potential (\(E^\circ\)) for the overall reaction is found.

Step-by-step solution

Identify the reactions and the corresponding \(E_{i}^\circ\) and \(n_{i}\)

Here are four half-cell reactions: \[\mathrm{H}_{5} \mathrm{IO}_{6} + \mathrm{H}^{+} + 2\mathrm{e}^- \longrightarrow \mathrm{IO}_{3}^- + 3\mathrm{H}_{2} \mathrm{O} \quad (E_1^\circ = 1.60V, n_1 = 2)\] \[\mathrm{IO}_{3}^- + 6\mathrm{H}^{+} + 5\mathrm{e}^- \longrightarrow \frac{1}{2}\mathrm{I}_{2} + 3\mathrm{H}_{2}\mathrm{O} \quad (E_2^\circ = 1.19V, n_2 = 5)\] \[2\mathrm{HIO}+2\mathrm{H}^{+} + 2\mathrm{e}^- \longrightarrow \mathrm{I}_{2} + 2\mathrm{H}_{2}\mathrm{O} \quad (E_3^\circ = 1.45V, n_3 = 2)\] \[\mathrm{I}_{2}+2\mathrm{e}^- \longrightarrow 2\mathrm{I}^- \quad (E_4^\circ = 0.535V, n_4 = 2)\]

Perform the calculations

Plugging in the values into the equation: \[E^\circ = \frac{\sum n_{i}E_{i}^\circ}{\sum n_{i}} = \frac{n_1E_1^\circ + n_2E_2^\circ + n_3E_3^\circ + n_4E_4^\circ}{n_1 + n_2 + n_3 + n_4}\] \[E^\circ = \frac{(2)(1.60V) + (5)(1.19V) + (2)(1.45V) + (2)(0.535V)}{2 + 5 + 2 + 2}\] Calculating this gives the value of \(E^\circ\).

Simplify the result

Simplify the expression above to get the final answer.

Key concepts

Half-Cell Reactions

In electrochemistry, a half-cell reaction plays a crucial role. A half-cell reaction involves either the reduction or oxidation process that occurs at one electrode. It can be thought of as half of the electrochemical cell's reaction. These reactions happen separately, but collectively they make up the entire cell reaction.
For example, consider the reaction of hydrogen peroxide to water, or iron (III) to iron (II) ions. In each half-cell, electrons are either gained or lost, hence responsible for generating an electrical potential.

• This electron exchange is crucial for the overall electrochemical process.
• The potential created by a half-cell reaction is known as the half-cell potential.
• Understanding half-cell reactions helps in determining the overall cell potential.

This concept is foundational in learning how to balance reactions in electrochemistry. It's almost like balancing a seesaw; each side contributes towards the equilibrium or a favorable cell potential.

Electrochemistry

Electrochemistry is the branch of chemistry that studies the interaction of electricity and chemical change. This process involves the movement of electrons between the electrodes in a solution of electrolytes. It bridges the concepts of chemical reactions and electrical flow.
The heart of electrochemistry is the electrochemical cell, which consists of two electrodes, the anode, and the cathode, submerged into an electrolyte solution. Each electrode involves a half-cell reaction.

• When an electrical circuit is completed, electrons flow from one electrode to another, triggering these reactions.
• This electron transfer results in an electric current, which can be harnessed to do work, like powering devices or plating metals.
• Electrochemistry is used in various applications such as batteries, fuel cells, corrosion prevention, and electrolysis.

It offers insights into processes like energy storage, chemical synthesis, and even new discoveries in science fields, making it a vital area of study.

Reduction Potential Calculation

Reduction potential, typically denoted as 𝑬°, measures an atom or molecule's ability to gain electrons. It's a fundamental part of predicting how easily a substance can be reduced in a reaction. To calculate the standard reduction potential of combined reactions that aren't directly observable, electrochemists use the formula:
\[E^{ ext{°}} = \frac{\sum n_{i} E_{i}^{ ext{°}}}{\sum n_{i}}\]
Here's how this works: for each half-cell reaction, determine both the standard reduction potential (\(E_{i}^\circ\)) and the number of electrons (n_{i}) involved. Then, multiply each standard reduction potential by its respective number of electrons, sum these results, and divide by the total number of electrons involved in all reactions.

• This method allows calculating a composite potential from multiple half-cells.
• It's based on the conservation and balance of electrons, ensuring accurate predictions of the cell potential.
• This formula especially aids when dealing with complex reactions where the potential isn't easily measurable.

Understanding reduction potential calculations helps in mastering electrochemical equations, assisting in designing efficient energy systems and laboratory setups.