Question

Calculate \(\mathscr{E}^{\circ}\) values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations that are not already balanced. Standard reduction potentials are found in Table \(18.1 .\) a. \(\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{H}^{-}(a q)\) b. \(\mathrm{Au}^{3+}(a q)+\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{Au}(s)\)

Short answer

Under standard conditions, the first cell involving hydrogen gas forming hydride and proton is not spontaneous, with a standard cell potential of \(\mathscr{E}^\circ = 0.00 \ V\). On the other hand, the second cell involving gold(III) being reduced in the presence of silver is spontaneous, with a standard cell potential of \(\mathscr{E}^\circ = 0.701 \ V\).

Step-by-step solution

Identify the half-reactions

For each cell, we need to separate the given reaction into the oxidation half-reaction and the reduction half-reaction. a. In the first case, the reaction is \(H_{2}(g) \longrightarrow H^{+}(aq) + H^{-}(aq)\) Here, hydrogen gas is being oxidized to form a proton and a hydride ion \(H^-\). So, we have two half reactions: Oxidation: \(H_2(g) \longrightarrow 2H^+(aq) + 2e^-\) Reduction: \(2e^- + 2H^+(aq) \longrightarrow H_2(g) + 2H^-\) b. In the second case, the reaction is \(Au^{3+}(aq) + Ag(s) \longrightarrow Ag^+(aq) + Au(s)\) Here, silver is being oxidized, and gold(III) is being reduced: Oxidation: \(Ag(s) \longrightarrow Ag^+(aq) + e^-\) Reduction: \(Au^{3+}(aq) + 3e^- \longrightarrow Au(s)\)

Find the standard reduction potentials

Now, we need to look up the standard reduction potentials for each half-reaction in Table 18.1: a. \(H_2(g)\) is the standard state, so its standard reduction potential is 0.00 V: \(2e^- + 2H^+(aq) \longrightarrow H_2(g)\), \(E^\circ = 0.00 \ V\) b. From Table 18.1, we get the following standard reduction potentials: \(Ag^+ (aq) + e^- \longrightarrow Ag(s)\), \(E^\circ = 0.799 \ V\) \(Au^{3+}(aq) + 3e^- \longrightarrow Au(s)\), \(E^\circ = 1.50 \ V\)

Calculate the cell potentials

We calculate the standard cell potential by adding the standard reduction potentials for the oxidation and reduction half-reactions: a. For the first cell, since the standard reduction potential for hydrogen is 0.00 V, the cell potential will also be 0.00 V: \(\mathscr{E}^\circ = 0.00 \ V\) b. For the second cell, the standard cell potential is given by: \(\mathscr{E}^\circ = E_{reduction}^\circ - E_{oxidation}^\circ\) Here, the reduction half-reaction is that of gold, and the oxidation half-reaction is that of silver. So, we have: \(\mathscr{E}^\circ = (1.50 \ V) - (0.799 \ V) = 0.701 \ V\)

Determine the spontaneity of the reactions

A reaction is spontaneous under standard conditions if its standard cell potential is positive: a. For the first cell, we have \(\mathscr{E}^\circ = 0.00 \ V\), which is not positive. Thus, the reaction is not spontaneous. b. For the second cell, we have \(\mathscr{E}^\circ = 0.701 \ V\), which is positive. Thus, the reaction is spontaneous.

Summary

Under standard conditions, the first cell (hydrogen gas forming hydride and proton) is not spontaneous, while the second cell (gold(III) being reduced in the presence of silver) is spontaneous. The standard cell potentials for these reactions are 0.00 V and 0.701 V, respectively.

Key concepts

Standard Reduction Potential

The standard reduction potential is a crucial concept in electrochemistry. It represents the tendency of a chemical species to gain electrons and be reduced. The standard reduction potential is measured in volts (\(V\)) under standard conditions, which means solutions are 1 M, gases are at 1 atm pressure, and the temperature is 25°C.

• In Table 18.1 of your textbook, you will find many standard reduction potentials listed for different half-reactions.
• A positive standard reduction potential indicates a strong tendency to gain electrons, while a negative value implies a weaker tendency.

When solving problems, knowledge of the standard reduction potential allows you to determine how likely a reaction is to occur, which direction it will progress in, and how cell potentials are calculated.

Spontaneous Reactions

For a reaction to be spontaneous under standard conditions, it must have a positive cell potential (\(\mathscr{E}^{\circ} > 0\)). Spontaneity means the reaction can proceed without any outside energy being added, and the process will tend to move in a forward direction naturally.

To determine if a reaction is spontaneous:

• Calculate the standard cell potential using standard reduction potentials from reference tables.
• If the calculated \(\mathscr{E}^{\circ}\) is positive, the reaction is spontaneous as written.
• If zero or negative, it indicates the reaction is not spontaneous in the written direction under standard conditions.

Thus, a positive standard cell potential is an indicator of a feasible electrochemical reaction under the given conditions.

Half-Reaction Balancing

In electrochemistry, balancing half-reactions is vital for accurately determining the overall reaction dynamics. Each half-reaction represents either oxidation or reduction, where electron transfer takes place.

• Oxidation involves the loss of electrons.
• Reduction entails the gain of electrons.

Balancing these half-reactions includes ensuring that both mass and charge are balanced:

• First, balance all non-oxygen and non-hydrogen atoms.
• Then, balance oxygen atoms by adding \(H_2O\) molecules where necessary.
• For hydrogen atoms, add \(H^+\) ions.
• Finally, balance the electrons so that the number of electrons lost in the oxidation process equals the number gained in the reduction process.

Perfectly balancing half-reactions is essential for correctly determining the overall cell potential and reaction spontaneity.

Cell Potential Calculation

The calculation of cell potential (\(\mathscr{E}^{\circ}\)) is a fundamental skill in electrochemistry for assessing the feasibility and spontaneity of reactions. It involves the use of standard reduction potentials from the table to predict the reaction’s capability to occur without external interference.

The formula to calculate the standard cell potential is:\[\mathscr{E}^{\circ} = E_{\text{reduction}}^{\circ} - E_{\text{oxidation}}^{\circ}\]Here’s how it works:

• Identify the half-reactions involved: one for oxidation and one for reduction.
• Use the standard reduction potentials from a reference table for each reaction.
• Subtract the oxidation potential from the reduction potential.

Each calculation gives an idea of how the electron flow will contribute to the overall cell voltage, determining if a process can spontaneously drive electrical work.