Determining Redox Reaction Spontaneity Using Standard Electrode Potentials

This question asks us to identify which redox reaction among the given options will NOT happen on its own (spontaneously) under standard conditions ($298 \, K$, $1 \, atm$ pressure, $1 \, M$ concentration). We can figure this out by looking at the standard electrode potentials ($E^\circ$) provided for different metal/ion pairs.

Understanding Spontaneity and Electrode Potentials

A redox reaction involves the transfer of electrons. One substance gets oxidized (loses electrons), and another gets reduced (gains electrons). The tendency for a substance to be reduced is measured by its standard reduction potential ($E^\circ$).

• A higher standard reduction potential indicates a greater tendency for the species to be reduced.
• A lower standard reduction potential indicates a greater tendency for the species to be oxidized (meaning its reverse reaction, oxidation, is more likely).

For a reaction to be spontaneous, the overall cell potential ($E^\circ_{cell}$) must be positive. We calculate $E^\circ_{cell}$ using the formula:
$ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} $ where:

• $E^\circ_{cathode}$ is the standard reduction potential of the species being reduced.
• $E^\circ_{anode}$ is the standard reduction potential of the species being oxidized.

Essentially, the species with the higher $E^\circ$ value acts as the cathode (gets reduced), and the species with the lower $E^\circ$ value acts as the anode (gets oxidized). If $E^\circ_{cell} > 0$, the reaction is spontaneous. If $E^\circ_{cell} < 0$, the reaction is non-spontaneous.

Given Standard Electrode Potentials

Let's list the provided standard electrode potentials at $298 \, K$:

Analyzing Each Reaction for Spontaneity

Option 1 Analysis: Copper and Iron Reaction

The reaction is: $Cu(s) + FeSO_4(aq) \rightarrow CuSO_4(aq) + Fe(s)$

In terms of ions, this is: $Cu(s) + Fe^{2+}(aq) \rightarrow Cu^{2+}(aq) + Fe(s)$

We need to determine which species is oxidized and which is reduced based on the potentials.

• Oxidation half-reaction: $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$
• Reduction half-reaction: $Fe^{2+}(aq) + 2e^- \rightarrow Fe(s)$

Comparing the standard reduction potentials:

• $E^\circ_{Cu^{2+}/Cu} = +0.34 \, V$
• $E^\circ_{Fe^{2+}/Fe} = -0.44 \, V$

The species with the higher reduction potential is $Cu^{2+}/Cu$ ($+0.34 \, V$), meaning $Cu^{2+}$ is more easily reduced. The species with the lower reduction potential is $Fe^{2+}/Fe$ ($-0.44 \, V$), meaning $Fe$ is more easily oxidized.

The reaction as written shows $Cu(s)$ being oxidized and $Fe^{2+}(aq)$ being reduced. This means $Cu$ is acting as the anode and $Fe^{2+}$ as the cathode.

Let's calculate the $E^\circ_{cell}$ for the given reaction using the standard reduction potentials:

The species being reduced ($Fe^{2+}$) indicates the cathode potential: $E^\circ_{cathode} = E^\circ_{Fe^{2+}/Fe} = -0.44 \, V$. The species being oxidized ($Cu$) requires its reduction potential value for the anode: $E^\circ_{anode} = E^\circ_{Cu^{2+}/Cu} = +0.34 \, V$.

Using the formula $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$:

$ E^\circ_{cell} = E^\circ_{Fe^{2+}/Fe} - E^\circ_{Cu^{2+}/Cu} $ $ E^\circ_{cell} = (-0.44 \, V) - (+0.34 \, V) $ $ E^\circ_{cell} = -0.44 \, V - 0.34 \, V $ $ E^\circ_{cell} = -0.78 \, V $

Since the calculated $E^\circ_{cell}$ is negative ($-0.78 \, V$), this reaction is NOT spontaneous under standard conditions. The reverse reaction, $Fe(s) + Cu^{2+}(aq) \rightarrow Fe^{2+}(aq) + Cu(s)$, would be spontaneous ($E^\circ_{cell} = +0.78 \, V$).

Option 2 Analysis: Zinc and Copper Reaction

The reaction is: $Zn(s) + Cu(NO_3)_2(aq) \rightarrow Zn(NO_3)_2(aq) + Cu(s)$

In terms of ions, this is: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

• Oxidation half-reaction: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$
• Reduction half-reaction: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$

Comparing the standard reduction potentials:

• $E^\circ_{Zn^{2+}/Zn} = -0.76 \, V$
• $E^\circ_{Cu^{2+}/Cu} = +0.34 \, V$

$Cu^{2+}$ ($+0.34 \, V$) has a higher reduction potential than $Zn^{2+}$ ($-0.76 \, V$). Therefore, $Cu^{2+}$ will be reduced (acting as the cathode), and $Zn$ will be oxidized (acting as the anode). This matches the reaction shown.

Calculating the $E^\circ_{cell}$:

$E^\circ_{cathode} = E^\circ_{Cu^{2+}/Cu} = +0.34 \, V$ $E^\circ_{anode} = E^\circ_{Zn^{2+}/Zn} = -0.76 \, V$

$ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} $ $ E^\circ_{cell} = (+0.34 \, V) - (-0.76 \, V) $ $ E^\circ_{cell} = +0.34 \, V + 0.76 \, V $ $ E^\circ_{cell} = +1.10 \, V $

Since the $E^\circ_{cell}$ is positive ($+1.10 \, V$), this reaction IS spontaneous under standard conditions.

Option 3 Analysis: Iron and Silver Reaction

The reaction is: $Fe(s) + 2AgNO_3(aq) \rightarrow Fe(NO_3)_2(aq) + 2Ag(s)$

In terms of ions, this is: $Fe(s) + 2Ag^{+}(aq) \rightarrow Fe^{2+}(aq) + 2Ag(s)$

• Oxidation half-reaction: $Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-$
• Reduction half-reaction: $2Ag^{+}(aq) + 2e^- \rightarrow 2Ag(s)$

Comparing the standard reduction potentials:

• $E^\circ_{Fe^{2+}/Fe} = -0.44 \, V$
• $E^\circ_{Ag^{+}/Ag} = +0.80 \, V$

$Ag^{+}$ ($+0.80 \, V$) has a higher reduction potential than $Fe^{2+}$ ($-0.44 \, V$). Thus, $Ag^{+}$ will be reduced (cathode), and $Fe$ will be oxidized (anode). This matches the reaction shown.

Calculating the $E^\circ_{cell}$:

$E^\circ_{cathode} = E^\circ_{Ag^{+}/Ag} = +0.80 \, V$ $E^\circ_{anode} = E^\circ_{Fe^{2+}/Fe} = -0.44 \, V$

$ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} $ $ E^\circ_{cell} = (+0.80 \, V) - (-0.44 \, V) $ $ E^\circ_{cell} = +0.80 \, V + 0.44 \, V $ $ E^\circ_{cell} = +1.24 \, V $

Since the $E^\circ_{cell}$ is positive ($+1.24 \, V$), this reaction IS spontaneous under standard conditions.

Option 4 Analysis: Zinc and Silver Reaction

The reaction is: $Zn(s) + 2AgNO_3(aq) \rightarrow Zn(NO_3)_2(aq) + 2Ag(s)$

In terms of ions, this is: $Zn(s) + 2Ag^{+}(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)$

• Oxidation half-reaction: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$
• Reduction half-reaction: $2Ag^{+}(aq) + 2e^- \rightarrow 2Ag(s)$

Comparing the standard reduction potentials:

• $E^\circ_{Zn^{2+}/Zn} = -0.76 \, V$
• $E^\circ_{Ag^{+}/Ag} = +0.80 \, V$

$Ag^{+}$ ($+0.80 \, V$) has a higher reduction potential than $Zn^{2+}$ ($-0.76 \, V$). Thus, $Ag^{+}$ will be reduced (cathode), and $Zn$ will be oxidized (anode). This matches the reaction shown.

Calculating the $E^\circ_{cell}$:

$E^\circ_{cathode} = E^\circ_{Ag^{+}/Ag} = +0.80 \, V$ $E^\circ_{anode} = E^\circ_{Zn^{2+}/Zn} = -0.76 \, V$

$ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} $ $ E^\circ_{cell} = (+0.80 \, V) - (-0.76 \, V) $ $ E^\circ_{cell} = +0.80 \, V + 0.76 \, V $ $ E^\circ_{cell} = +1.56 \, V $

Since the $E^\circ_{cell}$ is positive ($+1.56 \, V$), this reaction IS spontaneous under standard conditions.

Conclusion on Non-Spontaneous Reaction

After analyzing all the given reactions using their standard electrode potentials, we found that:

• Option 1 ($Cu(s) + Fe^{2+}(aq) \rightarrow Cu^{2+}(aq) + Fe(s)$) has a negative $E^\circ_{cell}$ ($-0.78 \, V$), indicating it is non-spontaneous.
• Option 2 ($Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$) has a positive $E^\circ_{cell}$ ($+1.10 \, V$), indicating it is spontaneous.
• Option 3 ($Fe(s) + 2Ag^{+}(aq) \rightarrow Fe^{2+}(aq) + 2Ag(s)$) has a positive $E^\circ_{cell}$ ($+1.24 \, V$), indicating it is spontaneous.
• Option 4 ($Zn(s) + 2Ag^{+}(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)$) has a positive $E^\circ_{cell}$ ($+1.56 \, V$), indicating it is spontaneous.

Therefore, the reaction that is NOT expected to occur spontaneously under standard conditions is the one presented in Option 1.