Problem 58 - Chemistry: An Atoms-Focused Approach

Question

What is the value of \(E^{\circ}\) for the following reaction? $$ 2 \mathrm{AgCl}(s)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{Ag}(s)+2 \mathrm{HCl}(a q) $$

Step-by-Step Solution

Answer

Answer: The standard cell potential (\(E^\circ\)) for the given reaction is \(+0.222~\text{V}\).

Step 1: Identify the half-reactions

We have to break down the given reaction into its oxidation and reduction half-reactions. Here, AgCl is being reduced to Ag, and H2 is being oxidized to H+ ions. Oxidation half-reaction: $$\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{H}^{+}(a q)+2e^{-}$$ Reduction half-reaction: $$2 \mathrm{AgCl}(s) + 2e^{-} \rightarrow 2 \mathrm{Ag}(s) + 2\mathrm{Cl}^{-}(a q)$$

Step 2: Find the standard reduction potentials for each half-reaction

Now we need to consult the standard reduction potentials table for both half-reactions: For the oxidation half-reaction, \(\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{H}^{+}(a q)+2e^{-}\), we read that the standard reduction potential (since H2 is a reference) is \(E^\circ = 0~\text{V}\). For the reduction half-reaction, \(\mathrm{AgCl}(s)+e^{-} \rightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q)\), we read that the standard reduction potential is \(E^\circ = +0.222~\text{V}\).

Step 3: Calculate the standard cell potential for the overall reaction

Since we have found the standard reduction potential for both half-reactions, we can now calculate the overall standard cell potential. The cell potential is the difference between the reduction and oxidation potentials: \(E^\circ_\text{cell} = E^\circ_\text{red} - E^\circ_\text{ox}\). In our case: \(E^\circ_\text{cell} = E^\circ_\text{red} - E^\circ_\text{ox} = (+0.222~\text{V}) - (0~\text{V}) = +0.222~\text{V}\) So the value of \(E^\circ\) for the given reaction is \(+0.222~\text{V}\).

Key Concepts

Understanding Oxidation Half-Reaction

The oxidation half-reaction is a crucial part of understanding redox (reduction-oxidation) reactions. During this process, a substance loses electrons. Losing electrons means that the substance is being oxidized.

In the given reaction, hydrogen gas \(\mathrm{H}_2(g)\) is oxidized to form hydrogen ions \(\mathrm{H}^+(aq)\). The oxidation half-reaction can be represented as:

• \(\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{H}^{+}(aq) + 2e^{-}\)

This transformation involves hydrogen gas losing electrons and becoming positive ions, known as protons. The electrons that are lost are then available to be gained by another substance, which we’ll see in the next section. It’s important to remember that oxidation always involves the loss of electrons.

Exploring Reduction Half-Reaction

In contrast to oxidation, a reduction half-reaction involves a substance gaining electrons. This causes the substance to become reduced. In the context of the given chemical equation, the substance being reduced is silver chloride \(\mathrm{AgCl}(s)\). It is reduced to silver metal \(\mathrm{Ag}(s)\) by gaining electrons.

The reduction half-reaction is written as:

• \(2 \mathrm{AgCl}(s) + 2e^{-} \rightarrow 2 \mathrm{Ag}(s) + 2\mathrm{Cl}^{-}(aq)\)

Here, each silver ion in the silver chloride compound gains an electron and is transformed into silver metal. Reduction is essentially the opposite of oxidation, involving the gain of electrons.

Understanding Standard Reduction Potentials

Standard reduction potentials are critical when analyzing electrochemical reactions. They allow us to predict the direction of electron flow and determine the feasibility of a redox reaction.

Each half-reaction has a specific standard reduction potential \(E^{\circ}\), measured in volts. In this problem:

• The standard reduction potential for hydrogen, which serves as a reference point, is \(0~\text{V}\).
• The reduction potential for \(\mathrm{AgCl}(s)\) transforming into \(\mathrm{Ag}(s)\) is \(+0.222~\text{V}\).

To find the overall standard cell potential, we use the formula:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{red}} - E^{\circ}_{\text{ox}} \] Here, the reduction potential is subtracted from the oxidation potential, resulting in the overall cell potential of \(+0.222~\text{V}\). This positive value indicates a spontaneous reaction under standard conditions.