Skip to main content
Sign in Create a free account
Varsity Tutors
Sign in

Cell Potential and Free Energy - AP Chemistry

  1. A galvanic cell under standard conditions uses the half-cells I2_2(s)|I−^-(aq) and Cl2_2(g)|Cl−^-(aq) with inert electrodes. The standard reduction potentials are: I2_2(s) + 2e−^- → 2I−^-(aq), E∘=+0.54 VE^\circ=+0.54\ \text{V}; Cl2_2(g) + 2e−^- → 2Cl−^-(aq), E∘=+1.36 VE^\circ=+1.36\ \text{V}. If the I2_2/I−^- half-cell is designated as the anode, which statement is correct about the cell reaction?
    Answer: The reaction is spontaneous and Ecell∘E^\circ_{\text{cell}} is positive. This question tests understanding of cell potential and free energy. With I₂/I⁻ as the anode (E° = +0.54 V) and Cl₂/Cl⁻ as the cathode (E° = +1.36 V), iodide is oxidized to I₂ and Cl₂ is reduced to Cl⁻. The standard cell potential is E°cell = E°cathode - E°anode = +1.36 V - (+0.54 V) = +0.82 V. Since E°cell is positive, the reaction is spontaneous under standard conditions. Students might incorrectly think that because both half-cells have positive reduction potentials, they cannot form a galvanic cell, but this misunderstands that the key is the difference in potentials. The strategy is to identify which species has the higher reduction potential (cathode) and calculate E°cell = E°cathode - E°anode.
  2. A student is given two half-reactions with their standard reduction potentials: - Sn2+^{2+}(aq) + 2e−^- → Sn(s) E∘=−0.14 VE^\circ = -0.14\,\text{V} - Br2_2(l) + 2e−^- → 2Br−^-(aq) E∘=+1.07 VE^\circ = +1.07\,\text{V} The student constructs a cell in which Sn(s) is the anode and the Br2_2/Br−^- half-cell is the cathode. Under standard conditions, which statement is correct?
    Answer: The cell reaction is spontaneous, and Ecell∘E^\circ_{\text{cell}} is positive. This question assesses the skill of cell potential and free energy. Half-cells with higher reduction potentials are more likely to be reduced, so they act as cathodes, while those with lower potentials are oxidized at the anode. The sign of E°cell, found by E°cathode - E°anode, determines spontaneity: positive for spontaneous reactions due to negative ΔG°. Here, Br2's much higher E° than Sn's leads to a positive E°cell and spontaneous cell reaction. A distractor like choice B errs by suggesting spontaneity with negative E°cell, based on the misconception that the negative E° for Sn overrides the positive difference. A transferable strategy is to calculate E°cell from relative potentials and confirm that a positive value signifies a spontaneous process in galvanic cells.
  3. A cell is built under standard conditions using the half-cells Cr(s)|Cr3+^{3+}(aq) and Co(s)|Co2+^{2+}(aq). Standard reduction potentials: Cr3+^{3+} + 3e−^- 6 Cr(s), E∘=−0.74 VE^\circ = -0.74\,\text{V}; Co2+^{2+} + 2e−^- 6 Co(s), E∘=−0.28 VE^\circ = -0.28\,\text{V}. If the Cr electrode is labeled as the cathode, which statement is correct about the cell as labeled?
    Answer: The reaction is nonspontaneous and Ecell∘E^\circ_{\text{cell}} is negative. This question assesses the skill of cell potential and free energy. With Cr labeled as cathode (-0.74 V) and Co as anode (-0.28 V), but Co has higher E°. E°cell = -0.74 V - (-0.28 V) = -0.46 V, negative. Negativity means nonspontaneous as labeled. A tempting distractor is choice A, which is incorrect because it assumes oxidation occurs at the electrode with the larger reduction potential, ignoring the labeled negative E°cell. Calculate based on labels; a positive E°cell indicates a spontaneous galvanic process.
  4. A student proposes a galvanic cell under standard conditions in which Mg(s) is the cathode and Cu(s) is the anode. Relevant standard reduction potentials: Mg2+^{2+} + 2e−^- 6 Mg(s), E∘=−2.37 VE^\circ = -2.37\,\text{V}; Cu2+^{2+} + 2e−^- 6 Cu(s), E∘=+0.34 VE^\circ = +0.34\,\text{V}. Based on these values, which statement is correct?
    Answer: The reaction is nonspontaneous and Ecell∘E^\circ_{\text{cell}} is negative. This question assesses the skill of cell potential and free energy. The proposed setup has Mg as cathode (-2.37 V) and Cu as anode (+0.34 V), but Cu has the higher reduction potential. Thus, E°cell = -2.37 V - (+0.34 V) = -2.71 V, which is negative. A negative E°cell means the reaction is nonspontaneous as proposed. A tempting distractor is choice A, which is incorrect because it assumes oxidation occurs at the electrode with the larger reduction potential, ignoring the labeled setup's negative E°cell. Always compute E°cell using the specified cathode and anode; a positive E°cell indicates a spontaneous galvanic process.
  5. Two standard reduction half-reactions are listed below: - F2_2(g) + 2e−^- → 2F−^-(aq) E∘=+2.87 VE^\circ = +2.87\,\text{V} - Au3+^{3+}(aq) + 3e−^- → Au(s) E∘=+1.50 VE^\circ = +1.50\,\text{V} A galvanic cell is constructed so that Au(s) is oxidized at the anode and F2_2 is reduced at the cathode. Under standard conditions, which statement is correct?
    Answer: The cell reaction is spontaneous, and Ecell∘E^\circ_{\text{cell}} is positive. This question assesses the skill of cell potential and free energy. The cathode is where the species with the highest reduction potential is reduced, and the anode is where oxidation occurs for the lower potential species. E°cell is positive when this natural order is followed, indicating spontaneity and a favorable free energy change. With F2's higher E° than Au's, the setup yields positive E°cell and spontaneous reaction. A distractor like choice B wrongly states spontaneity with negative E°cell, arising from the misconception that the highest E° should be oxidized at the anode. A transferable strategy is to select the higher reduction potential as cathode and ensure E°cell positivity denotes a spontaneous galvanic cell reaction.
  6. A student builds a standard electrochemical cell using a Ni(s)|Ni2+^{2+}(aq) half-cell and an Ag(s)|Ag+^+(aq) half-cell. The standard reduction potentials are: Ni2++2e−→Ni(s)\text{Ni}^{2+}+2e^-\rightarrow \text{Ni}(s), E∘=−0.25 VE^\circ=-0.25\,\text{V}; Ag++e−→Ag(s)\text{Ag}^+ + e^-\rightarrow \text{Ag}(s), E∘=+0.80 VE^\circ=+0.80\,\text{V}. If the cell operates as a galvanic cell, which statement is correct about spontaneity and the sign of Ecell∘E^\circ_{\text{cell}} for the overall reaction as written with Ag+^+ reduced?
    Answer: The reaction is spontaneous and Ecell∘E^\circ_{\text{cell}} is positive. This question examines cell potential and free energy relationships when Ag⁺ is reduced. Since Ag⁺ has a higher reduction potential (+0.80 V) than Ni²⁺ (-0.25 V), Ag⁺ will be reduced at the cathode while Ni is oxidized at the anode. The cell potential is E°cell = E°cathode - E°anode = (+0.80 V) - (-0.25 V) = +1.05 V. A positive E°cell corresponds to negative ΔG° and spontaneous reaction. Students might select option C by incorrectly thinking that a spontaneous reaction requires E°cell to be negative, confusing the signs of E°cell and ΔG°. The key principle is that positive E°cell always means spontaneous reaction in galvanic cells.
  7. A galvanic cell is constructed with a Zn(s)|Zn2+^{2+}(aq) half-cell and a Cu(s)|Cu2+^{2+}(aq) half-cell under standard conditions. A simplified table of standard reduction potentials is provided: Cu2+^{2+} + 2e−^- 6 Cu(s), E∘=+0.34 VE^\circ = +0.34\,\text{V}; Zn2+^{2+} + 2e−^- 6 Zn(s), E∘=−0.76 VE^\circ = -0.76\,\text{V}. If the cell operates as written (Zn electrode is the anode and Cu electrode is the cathode), which statement is correct about spontaneity and the sign of Ecell∘E^\circ_{\text{cell}}?
    Answer: The reaction is spontaneous and Ecell∘E^\circ_{\text{cell}} is positive. This question assesses the skill of cell potential and free energy. The standard cell potential, E°cell, is determined by subtracting the standard reduction potential of the anode from that of the cathode. Here, the Cu half-cell has a higher reduction potential (+0.34 V) compared to Zn (-0.76 V), so when Cu is the cathode and Zn is the anode, E°cell = +0.34 V - (-0.76 V) = +1.10 V, which is positive. A positive E°cell indicates that the redox reaction is spontaneous under standard conditions, as it corresponds to a negative ΔG°. A tempting distractor is choice D, which is incorrect because it assumes oxidation occurs at the electrode with the larger reduction potential, leading to a reversed and negative E°cell. Always identify the cathode as the half-cell with the higher reduction potential for spontaneous reactions; a positive E°cell indicates a spontaneous galvanic process.
  8. A standard cell is proposed using the half-reactions below: Cl2_2(g) + 2e−^- →\rightarrow 2Cl−^-(aq), E∘=+1.36 VE^\circ = +1.36\,\text{V}; Br2_2(l) + 2e−^- →\rightarrow 2Br−^-(aq), E∘=+1.07 VE^\circ = +1.07\,\text{V}. The overall reaction is written as Cl2_2(g) + 2Br−^-(aq) →\rightarrow 2Cl−^-(aq) + Br2_2(l). Which statement is correct about spontaneity and the sign of Ecell∘E^\circ_{\text{cell}} under standard conditions?
    Answer: The reaction is spontaneous and Ecell∘E^\circ_{\text{cell}} is positive. This question tests understanding of cell potential and free energy. In the reaction, Cl₂ is reduced to Cl⁻ (E° = +1.36 V) while Br⁻ is oxidized to Br₂ (reverse of the reduction, so we use E° = +1.07 V for Br₂/Br⁻). The cell potential is E°cell = E°cathode - E°anode = (+1.36 V) - (+1.07 V) = +0.29 V. Since E°cell is positive, the reaction is spontaneous. The misconception in option E overgeneralizes that halogens always oxidize halide ions, but the relative reduction potentials determine which halogen can oxidize which halide. The principle to remember is that the species with the higher reduction potential will oxidize the species with the lower reduction potential.
  9. An electrochemical cell under standard conditions uses a Mg(s) | Mg2+^{2+}(aq) half-cell and a H2_2(g) | H+^+(aq) half-cell (standard hydrogen electrode). The standard reduction potentials are: Mg2+^{2+} + 2e−^- →\rightarrow Mg(s) E∘=−2.37 VE^\circ = -2.37\,\text{V} 2H+^+(aq) + 2e−^- →\rightarrow H2_2(g) E∘=0.00 VE^\circ = 0.00\,\text{V} If the cell is arranged so that H2_2(g) is produced at the cathode, which statement is correct about the cell reaction?
    Answer: The reaction is spontaneous and Ecell∘E^\circ_{\text{cell}} is positive. This question examines cell potential and free energy using the standard hydrogen electrode. If H₂(g) is produced at the cathode, then reduction is occurring: 2H⁺ + 2e⁻ → H₂, making this the cathode reaction. By elimination, Mg must be oxidized at the anode: Mg → Mg²⁺ + 2e⁻. The standard cell potential is E°cell = E°cathode - E°anode = (0.00 V) - (-2.37 V) = +2.37 V. A positive E°cell indicates a spontaneous reaction. Students might incorrectly assume that because the hydrogen electrode has E° = 0, the cell potential must be zero, but this ignores the contribution from the other half-cell. Remember that the standard hydrogen electrode serves as a reference point (0.00 V) but still participates in calculating E°cell.
  10. A student connects two standard half-cells to make an electrochemical cell: Al(s) | Al3+^{3+}(aq) and Pb(s) | Pb2+^{2+}(aq). The standard reduction potentials are: Al3+^{3+} + 3e−^- →\rightarrow Al(s) E∘=−1.66 VE^\circ = -1.66\,\text{V} Pb2+^{2+} + 2e−^- →\rightarrow Pb(s) E∘=−0.13 VE^\circ = -0.13\,\text{V} If the overall cell reaction is written with Al(s) being oxidized and Pb2+^{2+}(aq) being reduced, which statement is correct?
    Answer: The reaction is spontaneous and Ecell∘E^\circ_{\text{cell}} is positive. This question tests cell potential and free energy calculations with different electron stoichiometries. The reaction as written has Al being oxidized (Al → Al³⁺ + 3e⁻) at the anode and Pb²⁺ being reduced (Pb²⁺ + 2e⁻ → Pb) at the cathode. The standard cell potential is E°cell = E°cathode - E°anode = (-0.13 V) - (-1.66 V) = +1.53 V. A positive E°cell indicates the reaction is spontaneous. A common error is to worry about balancing electrons (3e⁻ vs 2e⁻) when calculating E°cell, but standard potentials are intensive properties independent of stoichiometry. The strategy is to identify which species is reduced (cathode) and which is oxidized (anode), then apply E°cell = E°cathode - E°anode.
  11. A standard cell is made from two half-cells: Cd(s)|Cd2+^{2+}(aq) and Pb(s)|Pb2+^{2+}(aq). The standard reduction potentials are: Cd2+^{2+} + 2e−^- → Cd(s), E∘=−0.40 VE^\circ=-0.40\ \text{V}; Pb2+^{2+} + 2e−^- → Pb(s), E∘=−0.13 VE^\circ=-0.13\ \text{V}. If the Pb electrode is the cathode, which statement is correct about the cell reaction?
    Answer: The reaction is spontaneous and Ecell∘E^\circ_{\text{cell}} is positive. This question tests understanding of cell potential and free energy. With Pb as the cathode (E° = -0.13 V) and Cd as the anode (E° = -0.40 V), Pb²⁺ is reduced and Cd is oxidized, which makes sense because Pb²⁺/Pb has the less negative (more positive) reduction potential. The standard cell potential is E°cell = E°cathode - E°anode = -0.13 V - (-0.40 V) = +0.27 V. A positive E°cell indicates the reaction is spontaneous under standard conditions. A common misconception is assuming that negative reduction potentials for both electrodes mean the cell cannot be spontaneous, but this ignores the relative values. The strategy is to remember that the electrode with the more positive (or less negative) reduction potential serves as the cathode in a spontaneous galvanic cell.
  12. A student connects two half-cells under standard conditions: MnO4−_4^-(aq), Mn2+^{2+}(aq) in acidic solution with an inert electrode, and Sn(s)|Sn2+^{2+}(aq). Standard reduction potentials: MnO4−_4^- + 8H+^+ + 5e−^- 6 Mn2+^{2+} + 4H2_2O, E∘=+1.51 VE^\circ = +1.51\,\text{V}; Sn2+^{2+} + 2e−^- 6 Sn(s), E∘=−0.14 VE^\circ = -0.14\,\text{V}. If MnO4−_4^- is reduced at the cathode, which statement is correct?
    Answer: The reaction is spontaneous and Ecell∘E^\circ_{\text{cell}} is positive. This question assesses the skill of cell potential and free energy. MnO₄⁻ is reduced at the cathode (+1.51 V), making Sn the anode (-0.14 V). E°cell = +1.51 V - (-0.14 V) = +1.65 V, positive. This indicates spontaneity. A tempting distractor is choice A, which is incorrect because it assumes oxidation occurs at the electrode with the larger reduction potential, leading to confusion in sign and spontaneity. Verify the cathode has the higher E° for spontaneity; a positive E°cell indicates a spontaneous galvanic process.
  13. A galvanic cell is constructed using a Zn(s)|Zn2+^{2+}(aq) half-cell and a Cu(s)|Cu2+^{2+}(aq) half-cell under standard conditions. A simplified table of standard reduction potentials is provided: Zn2+^{2+}(aq) + 2e−^- →\rightarrow Zn(s), E∘=−0.76 VE^\circ = -0.76\,\text{V}; Cu2+^{2+}(aq) + 2e−^- →\rightarrow Cu(s), E∘=+0.34 VE^\circ = +0.34\,\text{V}. For the cell reaction written as Zn(s) + Cu2+^{2+}(aq) →\rightarrow Zn2+^{2+}(aq) + Cu(s), which statement is correct about spontaneity and the sign of Ecell∘E^\circ_{\text{cell}}?
    Answer: The reaction is spontaneous and Ecell∘E^\circ_{\text{cell}} is positive. This question tests understanding of cell potential and free energy. In a galvanic cell, the species with the more positive reduction potential (Cu²⁺/Cu at +0.34 V) undergoes reduction at the cathode, while the species with the more negative reduction potential (Zn²⁺/Zn at -0.76 V) undergoes oxidation at the anode. The standard cell potential is calculated as E°cell = E°cathode - E°anode = (+0.34 V) - (-0.76 V) = +1.10 V. Since E°cell is positive, the reaction is spontaneous under standard conditions. A common misconception is thinking that a negative reduction potential for one half-cell makes the overall reaction nonspontaneous, but what matters is the difference between the two potentials. Remember that for galvanic cells, a positive E°cell always indicates a spontaneous reaction with negative ΔG°.
  14. A galvanic cell is built under standard conditions from a Mg(s)|Mg2+^{2+}(aq) electrode and a Cd(s)|Cd2+^{2+}(aq) electrode. The standard reduction potentials are: Mg2+^{2+} + 2e−^- →\rightarrow Mg(s), E∘=−2.37 VE^\circ = -2.37\,\text{V}; Cd2+^{2+} + 2e−^- →\rightarrow Cd(s), E∘=−0.40 VE^\circ = -0.40\,\text{V}. The student labels Cd as the anode and Mg as the cathode. Based on the provided E∘E^\circ values, which statement is correct about the cell reaction as labeled?
    Answer: The reaction is nonspontaneous and Ecell∘E^\circ_{\text{cell}} is negative. This question tests understanding of cell potential and free energy. The student labeled Cd as anode (oxidation) and Mg as cathode (reduction), meaning the reaction would be Cd → Cd²⁺ + 2e⁻ and Mg²⁺ + 2e⁻ → Mg. This gives E°cell = E°cathode - E°anode = (-2.37 V) - (-0.40 V) = -1.97 V. A negative E°cell indicates the reaction is nonspontaneous as labeled. The error in option E is thinking that having the more negative reduction potential automatically makes Mg the anode, but the student's labeling forces Mg to be the cathode, creating a nonspontaneous cell. The key lesson is that incorrect electrode labeling can result in a nonspontaneous cell with negative E°cell.
  15. A student compares two standard reduction potentials to predict whether a proposed galvanic cell reaction is spontaneous: Cr3+^{3+} + 3e−^- →\rightarrow Cr(s) E∘=−0.74 VE^\circ = -0.74\,\text{V} Co2+^{2+} + 2e−^- →\rightarrow Co(s) E∘=−0.28 VE^\circ = -0.28\,\text{V} The student proposes a cell in which Co(s) is oxidized and Cr3+^{3+}(aq) is reduced. Which statement is correct about the proposed cell reaction under standard conditions?
    Answer: The reaction is nonspontaneous and Ecell∘E^\circ_{\text{cell}} is negative. This question tests understanding of cell potential and free energy for a proposed reaction. The student proposes Co being oxidized (Co → Co²⁺ + 2e⁻) at the anode and Cr³⁺ being reduced (Cr³⁺ + 3e⁻ → Cr) at the cathode. The standard cell potential is E°cell = E°cathode - E°anode = (-0.74 V) - (-0.28 V) = -0.46 V. A negative E°cell indicates the reaction as proposed is nonspontaneous. Students might incorrectly think that because Cr³⁺/Cr has a more negative reduction potential, it should be oxidized, but the question asks about the feasibility of the reaction as written. The strategy is to calculate E°cell for the proposed reaction direction; if negative, the reaction is nonspontaneous and would require external energy input.
  16. A standard galvanic cell is made from a Cr(s)|Cr3+^{3+}(aq) half-cell and a Br2_2(l)|Br−^-(aq) half-cell with an inert Pt electrode for the bromine half-cell. The standard reduction potentials are: Cr3++3e−→Cr(s)\text{Cr}^{3+}+3e^-\rightarrow \text{Cr}(s), E∘=−0.74 VE^\circ=-0.74\,\text{V}; Br2(l)+2e−→2Br−(aq)\text{Br}_2(l)+2e^-\rightarrow 2\text{Br}^-(aq), E∘=+1.07 VE^\circ=+1.07\,\text{V}. Which statement is correct for the cell reaction in which Br2_2 is reduced?
    Answer: The reaction is spontaneous and Ecell∘E^\circ_{\text{cell}} is positive. This problem examines cell potential and free energy in a Cr/Br₂ galvanic cell. With Br₂ having a much higher reduction potential (+1.07 V) than Cr³⁺ (-0.74 V), Br₂ is reduced at the cathode while Cr is oxidized at the anode. The cell potential is E°cell = E°cathode - E°anode = (+1.07 V) - (-0.74 V) = +1.81 V. A positive E°cell indicates negative ΔG° and spontaneous reaction. Students might choose option B by confusing the relationship between E°cell and spontaneity, thinking positive E°cell means nonspontaneous. Remember that for galvanic cells, positive E°cell always indicates spontaneous electron flow and energy release.
  17. A student builds a cell using Fe(s)|Fe2+^{2+}(aq) and Ag(s)|Ag+^+(aq) half-cells under standard conditions. Standard reduction potentials: Ag+^+ + e−^- 6 Ag(s), E∘=+0.80 VE^\circ = +0.80\,\text{V}; Fe2+^{2+} + 2e−^- 6 Fe(s), E∘=−0.44 VE^\circ = -0.44\,\text{V}. If electrons flow through the wire from the Fe electrode to the Ag electrode, which statement is correct?
    Answer: The reaction is spontaneous and Ecell∘E^\circ_{\text{cell}} is positive. This question assesses the skill of cell potential and free energy. Electron flow from Fe to Ag indicates Fe is the anode and Ag is the cathode, with Ag having a higher reduction potential (+0.80 V) than Fe (-0.44 V). Thus, E°cell = +0.80 V - (-0.44 V) = +1.24 V, which is positive. This positive value confirms the reaction is spontaneous, aligning with the direction of electron flow. A tempting distractor is choice A, which is incorrect because it assumes oxidation occurs at the electrode with the larger reduction potential, resulting in a negative E°cell and nonspontaneity. Always calculate E°cell based on the given electron flow direction; a positive E°cell indicates a spontaneous galvanic process.
  18. A standard galvanic cell is built from the half-cells Co(s)|Co2+^{2+}(aq) and Br2_2(l)|Br−^-(aq) with inert electrodes as needed. The standard reduction potentials are: Co2+^{2+} + 2e−^- → Co(s), E∘=−0.28 VE^\circ=-0.28\ \text{V}; Br2_2(l) + 2e−^- → 2Br−^-(aq), E∘=+1.07 VE^\circ=+1.07\ \text{V}. If Co(s) is the anode, which statement is correct about the overall cell reaction?
    Answer: The reaction is spontaneous and Ecell∘E^\circ_{\text{cell}} is positive. This question tests understanding of cell potential and free energy. With Co(s) as the anode, Co is oxidized to Co²⁺, and at the cathode, Br₂ is reduced to Br⁻. Since Br₂/Br⁻ has a much more positive reduction potential (+1.07 V) than Co²⁺/Co (-0.28 V), this assignment makes sense for a spontaneous reaction. The standard cell potential is E°cell = E°cathode - E°anode = +1.07 V - (-0.28 V) = +1.35 V. A positive E°cell confirms the reaction is spontaneous. A common misconception is thinking that the presence of a negative reduction potential at the anode makes the overall reaction nonspontaneous, but this ignores the large positive contribution from the cathode. The key principle is that a positive E°cell always indicates a spontaneous galvanic process.
  19. A standard electrochemical cell is constructed with a Ni(s)|Ni2+^{2+}(aq) half-cell and a Pb(s)|Pb2+^{2+}(aq) half-cell. The standard reduction potentials are: Ni2+^{2+} + 2e−^- →\rightarrow Ni(s), E∘=−0.25 VE^\circ = -0.25\,\text{V}; Pb2+^{2+} + 2e−^- →\rightarrow Pb(s), E∘=−0.13 VE^\circ = -0.13\,\text{V}. The overall reaction is written as Ni(s) + Pb2+^{2+}(aq) →\rightarrow Ni2+^{2+}(aq) + Pb(s). Which statement is correct about spontaneity and the sign of Ecell∘E^\circ_{\text{cell}}?
    Answer: The reaction is spontaneous and Ecell∘E^\circ_{\text{cell}} is positive. This question tests understanding of cell potential and free energy. In the reaction, Ni is oxidized (reverse of Ni²⁺ + 2e⁻ → Ni) and Pb²⁺ is reduced to Pb. Since Pb²⁺/Pb has E° = -0.13 V and Ni²⁺/Ni has E° = -0.25 V, the cell potential is E°cell = E°cathode - E°anode = (-0.13 V) - (-0.25 V) = +0.12 V. A positive E°cell indicates the reaction is spontaneous. The misconception in option E assumes that because both reduction potentials are negative, the reaction must be spontaneous, but what matters is which species has the more positive (less negative) reduction potential. Remember that even with two negative reduction potentials, the cell can be spontaneous if E°cell is positive.
  20. A student connects a Fe(s)|Fe2+^{2+}(aq) half-cell to a Ag(s)|Ag+^+(aq) half-cell under standard conditions. The standard reduction potentials are: Fe2+^{2+} + 2e−^- → Fe(s), E∘=−0.44 VE^\circ=-0.44\ \text{V}; Ag+^+ + e−^- → Ag(s), E∘=+0.80 VE^\circ=+0.80\ \text{V}. If electrons flow through the external circuit from the Fe electrode to the Ag electrode, which statement is correct about the overall cell reaction?
    Answer: The reaction is spontaneous and Ecell∘E^\circ_{\text{cell}} is positive. This question tests understanding of cell potential and free energy. When electrons flow from Fe to Ag through the external circuit, Fe is being oxidized (anode) and Ag⁺ is being reduced (cathode), which makes sense because Ag⁺/Ag has the more positive reduction potential (+0.80 V) compared to Fe²⁺/Fe (-0.44 V). The standard cell potential is E°cell = E°cathode - E°anode = +0.80 V - (-0.44 V) = +1.24 V. A positive E°cell indicates the reaction is spontaneous under standard conditions. Students might incorrectly think that because Fe has a negative reduction potential, the overall reaction must be nonspontaneous, but this confuses the individual electrode potential with the cell potential. The key strategy is to identify which electrode serves as cathode (more positive E°) and calculate E°cell = E°cathode - E°anode.

Start this quiz