Electrochemistry: Determine if a Reaction is Spontaneous
Question
Use data from the standard electrode reduction potentials table to predict whether the reaction below is spontaneous in the forward reaction (all reactants and products are in their standard states):
[latex]2\mathrm{Al}\,\mathrm{(s)} + 3\mathrm{Zn}^{2+}\mathrm{(aq)}\longrightarrow 2\mathrm{Al}^{3+}\mathrm{(aq)} + 3\mathrm{Zn}\,\mathrm{(s)}[/latex]
Show/Hide Answer
The reaction is spontaneous in the forward direction.
Refer to:
- Section 9.3: Standard Reduction Potentials (1).
- Table P1: Standard Reduction Potentials by Element (2).
Strategy Map
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Check out the strategy map.
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1. Using the provided reaction, identify which reactant species is being oxidized and which is being reduced.
Show/Hide Hint“LEO says GER”:
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2. Write out the half-reactions and balance them.
Show/Hide HintFirst, ensure that the masses are balanced using stoichiometric coefficients. Then, use electrons to balance the charge between the reactant and product side. Lost electrons are shown as products, while gained electrons are shown as reactants. |
3. Identify where each reaction occurs (at the anode/cathode).
Show/Hide Hint“An Ox” and “Red Cat”:
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4. Use Table P1: Standard Reduction Potentials by Element (2) or Table 9.3.1 (1) to look up the E° for each reaction.
Show/Hide HintBe sure to match the physical states in the table! |
5. Use the E°cell equation to calculate the E°cell for the total reaction.
Show/Hide Hint[latex]\mathrm{E}^\circ\text{cell}=\mathrm{E}^\circ\text{cathode}-\mathrm{E}^\circ\text{anode}[/latex] Subtract the reduction potential at the anode from the reduction potential at the cathode. Note: SI units for cell potential is volts. |
6. Use the answer from step 5 to determine if the reaction will be spontaneous or not.
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Solution
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Check out this solution.
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[latex]2\mathrm{Al}\,\mathrm{(s)}+3\mathrm{Zn}^{2+}\mathrm{(aq)}\longrightarrow2\mathrm{Al}^{3+}\mathrm{(aq)}+3\mathrm{Zn}\,\mathrm{(s)}[/latex]
Oxidation half reaction:
[latex](\mathrm{Al}\,\mathrm{(s)}\longrightarrow\mathrm{Al}^{3+}\mathrm{(aq)}+3\mathrm{e}^-)\times2[/latex]
Reduction half reaction:
[latex](\mathrm{Zn}^{2+}\mathrm{(aq)}+2\mathrm{e}^-\longrightarrow\mathrm{Zn}\,\mathrm{(s)})\times3[/latex]
[latex]\begin{aligned} \mathrm{E}^\circ\text{cell}&=\mathrm{E}^\circ\text{cathode}-\mathrm{E}^\circ\text{anode}\\ \mathrm{E}^\circ\text{cell}&=-0.7618\mathrm{~V}-(-1.676\mathrm{~V})=0.9139\mathrm{~V} \end{aligned}[/latex]
The cell potential is positive, so the reaction is spontaneous.
Answer: The reaction is spontaneous in the forward direction.
Guided Solution
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The guided solution below will give you the reasoning for each step to get your answer, with reminders and hints.
Show/Hide Guided Solution
| Guided Solution Ideas |
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| This question is a calculation type problem in which you must identify the cathode (reduction) and anode (oxidation) reactions to calculate the E°cell potential for the overall reaction. The sign of the value will allow you to predict if the reaction is spontaneous or not.
[latex]2\mathrm{Al}\,\mathrm{(s)}+3\mathrm{Zn}^{2+}\mathrm{(aq)}\longrightarrow2\mathrm{Al}^{3+}\mathrm{(aq)}+3\mathrm{Zn}\,\mathrm{(s)}[/latex] Show/Hide ResourcesRefer to: |
| Identify which reactant was oxidized and which was reduced.
Losing electrons; product more positively charged: [latex]\mathrm{Al}\,\mathrm{(s)}\longrightarrow\mathrm{Al}^{3+}\mathrm{(aq)}[/latex] Gaining electrons; product more negatively charged: [latex]\mathrm{Zn}^{2+}\mathrm{(aq)}\longrightarrow\mathrm{Zn}\,\mathrm{(s)}[/latex]
Show/Hide Don’t Forget!
You can remember this by thinking the phrase “LEO says GER” (Lose-electrons-oxidized) and (Gains-electrons-reduced). |
| Write out and balance the half-reactions to find their E° values.
Oxidation half-reaction (at anode as written): [latex]\mathrm{Al}\,\mathrm{(s)}\longrightarrow\mathrm{Al}^{3+}\mathrm{(aq)}+3\mathrm{e}^-[/latex] Reduction half-reaction (at cathode as written): [latex]\mathrm{Zn}^{2+}\mathrm{(aq)}+2\mathrm{e}^-\longrightarrow\mathrm{Zn}\,\mathrm{(s)}[/latex] Show/Hide Think About This!
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Recall the E°cell equation.
Show/Hide Don’t Forget![latex]\mathrm{E}^\circ\text{cell}=\mathrm{E}^\circ\text{cathode}-\mathrm{E}^\circ\text{anode}[/latex] You will look up the E° values for the half-reaction in Table P1: Standard Reduction Potentials by Element (2) to use in the equation. Standard reduction potential values: At anode: [latex]\mathrm{Al}^{3+}\mathrm{(aq)}+3\mathrm{e}^-\longrightarrow\mathrm{Al}\,\mathrm{(s)}[/latex] [latex]\mathrm{E}^\circ\text{reduction potential}=-1.676\mathrm{~V}[/latex] At cathode: [latex]\mathrm{Zn}^{2+}\mathrm{(aq)}+2\mathrm{e}^-\longrightarrow\mathrm{Zn}\,\mathrm{(s)}[/latex] [latex]\mathrm{E}^\circ\text{reduction potential}=-0.7618\mathrm{~V}[/latex] Show/Hide Think About This!Notice that the reactions in the standard reduction potential table are all written as reduction half-reactions; electrons are gained as reactants. Reversing the reactions gives the oxidation half-reaction. To determine the oxidation potential, simply change the sign of the reduction potential value. How does stoichiometry impact the calculation? Show/Hide Watch Out!The stoichiometry is not used to calculate the E° |
| The calculated cell potential will tell you if the reaction will be spontaneous or not:
[latex]\begin{aligned} \mathrm{E}^\circ\text{cell}&=\mathrm{E}^\circ\text{cathode}-\mathrm{E}^\circ\text{anode}\\ \mathrm{E}^\circ\text{cell}&=-0.7618\mathrm{~V}-(-1.676\mathrm{~V})=0.9139\mathrm{~V} \end{aligned}[/latex] Show/Hide Think About This!
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| Complete Solution |
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[latex]2\mathrm{Al}\,\mathrm{(s)}+3\mathrm{Zn}^{2+}\mathrm{(aq)}\longrightarrow2\mathrm{Al}^{3+}\mathrm{(aq)}+3\mathrm{Zn}\,\mathrm{(s)}[/latex] Oxidation half-reaction: [latex](\mathrm{Al}\,\mathrm{(s)}\longrightarrow\mathrm{Al}^{3+}\mathrm{(aq)}+3\mathrm{e}^-)\times2[/latex] The reaction is multiplied by 2 to balance the 2 equations. The electrons must cancel each other out before the half-reactions are added together. [latex]\mathrm{Al}\,\mathrm{(s)}\longrightarrow\mathrm{Al}^{3+}\mathrm{(aq)}+3\mathrm{e}^-[/latex] [latex]\mathrm{E}^\circ=-1.676\mathrm{~V}[/latex] Reduction half-reaction: [latex](\mathrm{Zn}^{2+}\mathrm{(aq)}+2\mathrm{e}^-\longrightarrow\mathrm{Zn}\,\mathrm{(s)})\times3[/latex] The reaction is multiplied by 3 to balance the number of electrons gained and lost between the 2 half-reactions. [latex]\mathrm{Zn}^{2+}\mathrm{(aq)}+2\mathrm{e}^-\longrightarrow\mathrm{Zn}\,\mathrm{(s)}[/latex] [latex]\mathrm{E}^\circ=-0.7618\mathrm{~V}[/latex] |
| Recall the standard cell potential equation:
[latex]\mathrm{E}^\circ\text{cell}=\mathrm{E}^\circ\text{cathode}-\mathrm{E}^\circ\text{anode}[/latex] Plug in the values from the provided table: [latex]\mathrm{E}^\circ\text{cell}=-0.7618\mathrm{~V}-(-1.676\mathrm{~V})=+0.9139\mathrm{~V}[/latex] The cell potential is positive, meaning the electrons are flowing away from the negatively charged anode and towards the positively charged cathode. Therefore, the reaction is spontaneous. Answer: The reaction is spontaneous in the forward direction. |
Check Your Work
Check to make sure that the species in the provided overall reaction that describes the reduction matches the reduction half-reaction details with the most positive standard reduction potential.
When the standard reduction potential at the cathode is more positive than the standard reduction potential at the anode, the cell potential will be positive and the reaction will be spontaneous.
Show/Hide Think About This!
[latex]2\mathrm{Al(s)}+3\boldsymbol{\mathrm{Zn}^{2+}\mathrm{(aq)}}\longrightarrow2\mathrm{Al}^{3+}\mathrm{(aq)}+3\boldsymbol{\mathrm{Zn(s)}}[/latex]
At anode:
[latex]\mathrm{Al}^{3+}\mathrm{(aq)}+3\mathrm{e}^-\longrightarrow\mathrm{Al(s)}[/latex]
[latex]\mathrm{E}^\circ\text{reduction potential}=-1.676\mathrm{~V}[/latex]
At cathode:
[latex]\boldsymbol{\mathrm{Zn}^{2+}\mathrm{(aq)}}+2\mathrm{e}^-\longrightarrow\boldsymbol{\mathrm{Zn(s)}}[/latex]
[latex]\mathrm{E}^\circ\text{reduction potential}=\boldsymbol{-0.7618\mathrm{~V}}[/latex] (more positive)
Refer to Table P1: Standard Reduction Potentials by Element (2).
Does your answer make chemical sense?
Show/Hide Answer
In electrochemistry, if there is a difference in half-cell reduction potentials, electrons will flow spontaneously away from the negatively charged anode towards the positively charged cathode.
In this type of question, we are determining if the overall reaction equation (with a single direction arrow) has been written correctly. The overall reaction equation defines the anode and cathode based on the species shown as being oxidized and reduced, respectively.
PASS Attribution
- LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (4).
- Question 9.E.8 (b) from LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (5) is used under a CC BY-NC-SA 4.0 license.
- Question 9.E.8 (b) is question Q11 (b) from LibreTexts Chem 1202 (6), which is under a CC BY-NC-SA 4.0 license.
References
1. OpenStax. 9.3: Standard Reduction Potentials. In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2022. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/09%3A_Electrochemistry/9.03%3A_Standard_Reduction_Potentials.
2. LibreTexts. P1: Standard Reduction Potentials by Element (Table). In Reference Tables; LibreTexts, 2021. https://chem.libretexts.org/Ancillary_Materials/Reference/Reference_Tables/Electrochemistry_Tables/P1%3A_Standard_Reduction_Potentials_by_Element.
3. OpenStax. 9.1: Balancing Oxidation-Reduction Reactions. In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2022. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/09%3A_Electrochemistry/9.01%3A_Balancing_Oxidation-Reduction_Reactions.
4. Thompson Rivers University. TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2024. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520).
5. Thompson Rivers University. 9.E: Exercises on Electrochemistry. In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2022. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/09%3A_Electrochemistry/9.E%3A_Exercises_on_Electrochemistry.
6. Mount Royal University. 6.9: Exercises on Electrochemistry. In Mount Royal University Chem1202; LibreTexts, (2021). https://chem.libretexts.org/Courses/Mount_Royal_University/Chem_1202/Unit_6%3A_Electrochemistry/6.9%3A_Exercises_on_Electrochemistry
