15 Electrochemistry — Identify Anode and Cathode and Calculate Cell Potential
Question
Identify where each half reaction will occur (on the cathode or anode), write the overall balanced reaction, and calculate the E∘ cell.
[latex]\begin{equation} \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) \end{equation}[/latex]
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[latex]\begin{equation} \begin{aligned} & \mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq}) \text { occurs at the cathode. } \\ & \mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \text {occurs at the anode. } \\ & \mathrm{E}^{\circ} \text { cell }=2.167 \mathrm{~V} \end{aligned} \end{equation}[/latex]
Refer to Section 9.3: Standard Reduction Potentials (1).
Data is from Table 9.3.1: Selected Standard Reduction Potentials at 25 °C (1) and P1: Standard Reduction Potentials by Element (Table) (2).
Strategy Map
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| Strategy Map Steps |
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1. Identify what is being oxidized and what is being reduced.
Show/Hide HintLook at the change in charge for each species. |
2. Write out the half reactions and balance them.
Show/Hide HintBalance mass and balance charge. |
3. Identify where each reaction occurs (at the anode/cathode).
Show/Hide HintThe oxidization reaction will occur at the anode, and the reduction reaction will occur at the cathode. |
| 4. Use the table in P1: Standard Reduction Potentials by Element (2) or Table 9.3.1: Selected Standard Reduction Potentials at 25 °C (1) to look up the E° for each reaction. |
5. Use the E°cell equation to calculate the E°cell for the total reaction.
Show/Hide HintE°cell = E°cathode – E°anode |
Solution
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Fe2+(aq) + Cl2(g) → Fe3+(aq) + 2Cl–(aq)
Reduction Reaction (at cathode):
[latex]\begin{equation} \begin{aligned} & \mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq}) \\ & \mathrm{E}^{\circ}=1.269 \mathrm{~V} \end{aligned} \end{equation}[/latex]
Oxidation Reaction (at anode):
[latex]\begin{equation} \begin{aligned} & \left(\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-}\right) \times 2 \\ & \mathrm{E}^{\circ}=-0.771 \mathrm{~V} \end{aligned} \end{equation}[/latex]
Overall Reaction
[latex]\begin{equation} 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) \end{equation}[/latex]
[latex]\begin{equation} \begin{aligned} & \mathrm{E}^{\circ} \text { cell }=\mathrm{E}^{\circ} \text { cathode }-\mathrm{E}^{\circ} \text { anode } \\ & \mathrm{E}^{\circ} \text { cell }=(1.396 \mathrm{~V})-(-0.771 \mathrm{~V})=2.167 \mathrm{~V} \end{aligned} \end{equation}[/latex]
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.
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| Guided Solution Ideas |
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This question is a calculation problem in which you must identify the cathode and anode reactions to calculate the E°cell potential for the overall reaction.
Show/Hide ResourceRefer to Section 9.3: Standard Reduction Potentials (1). |
You need to identify which reactant was oxidized and which reactant was reduced.
Show/Hide Think About This!
You can remember this by thinking of the phrase “LEO says GER” (Lose-electrons-oxidized) and (Gains-electrons-reduced). |
You need to write out and balance the half reactions to find their E° values.
Show/Hide Think About This!
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Which reaction occurs at the anode? Which reaction occurs at the cathode?
Show/Hide Think About This!The oxidization reaction will occur at the anode, and the reduction reaction will occur at the cathode. Think “An-Ox” (Anode-Oxidized) and “Red-Cat” (Reduction-Cathode). |
| Recall the E°cell equation.
You will look up the E° values for the half-reaction in the table in P1: Standard Reduction Potentials by Element (2) or Table 9.3.1: Selected Standard Reduction Potentials at 25 °C (1) and plug those values into the equation. What about the stoichiometry of the reactions? Show/Hide Watch Out!E°cell = E°cathode – E°anode Notice that some reactions in the table are written as the reduction reaction and some as the oxidization reaction. If you find the value for the opposite reaction than the one you need, flip the sign of the value. The stoichiometry does not impact the calculation of E°. |
| Complete Solution |
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| Fe2+(aq) + Cl2 (g) → Fe3+(aq) + 2Cl–(aq)
Iron is oxidized and chlorine is reduced. Reduction Reaction (occurs at cathode): [latex]\begin{equation} \begin{aligned} & \mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq}) \\ & \mathrm{E}^{\circ}=1.269 \mathrm{~V} \end{aligned} \end{equation}[/latex] Oxidation Reaction (occurs at anode): [latex]\begin{equation} \left(\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-}\right) \times 2 \end{equation}[/latex] Multiplied by two to make the electrons gained an lost equal so when the half-reactions are added together the electrons cancel each other out. [latex]\begin{equation} \begin{aligned} & 2 \mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow 2 \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{e}^{-} \\ & \mathrm{E}^{\circ}=-0.771 \mathrm{~V} \end{aligned} \end{equation}[/latex] Overall Reaction [latex]\begin{equation} 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) \end{equation}[/latex] |
| Recall the E°cell equation:
[latex]\begin{equation} \mathrm{E}^{\circ} \text { cell }=\mathrm{E}^{\circ} \text { cathode }-\mathrm{E}^{\circ} \text { anode } \end{equation}[/latex] Plug in the values from the table: [latex]\begin{equation} \mathrm{E}^{\circ} \mathrm{cell}=(1.396 \mathrm{~V})-(-0.771 \mathrm{~V})=\mathbf{2 . 1 6 7 V} \end{equation}[/latex] |
Check Your Work
You should balance the overall reaction in mass and the number of electrons lost and gained. The overall cell potential is a positive value, which means it is a spontaneous reaction.
Refer to Section 9.3: Standard Reduction Potentials (1).
Does your answer make chemical sense?
Show/Hide Answer
The cell potential (E°cell) is the potential difference between the two half-cells in an electrochemical cell. We can use this to calculate the likelihood of a reaction to occur spontaneously.
We use standard reduction potentials to calculate the cell potential, so the anode is subtracted to indicate its half reaction is reversed (occurs as the oxidation).
PASS Attribution
- LibreTexts PASS Chemistry Book CHEM 1510/1520 (3).
- Question 9.E.6(d) from LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (4) is used under a CC BY-NC-SA 4.0 license.
- Question 9.E.6(d) is question Q22(d) from LibreTexts Chem 1202 (5), 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. In Reference Tables. LibreTexts, 2021. https://chem.libretexts.org/Ancillary_Materials/Reference/Reference_Tables/Electrochemistry_Tables/P1%3A_Standard_Reduction_Potentials_by_Element.
3. Blackstock, L.; Brewer, S.; Jensen, A. PASS Chemistry Book CHEM 1510/1520; LibreTexts, 2023. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1510%2F%2F1520.
4. 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.
5. Mount Royal University. 6.9: Exercises on Electrochemistry. In Chem 1202. LibreTexts, 2021. https://chem.libretexts.org/Courses/Mount_Royal_University/Chem_1202/Unit_6%3A_Electrochemistry/6.9%3A_Exercises_on_Electrochemistry.
