Electrochemical Cells and Redox Reactions (4C) - Chem & Physics

In a conceptual prediction about cell reversal, a galvanic cell initially uses: Zn(s)|Zn $^{2+}$ (1.0 M) || Cu $^{2+}$ (1.0 M)|Cu(s). Later, Zn $^{2+}$ becomes very high and Cu $^{2+}$ becomes very low due to extended operation. Which conclusion is most consistent with the electrochemical process as these concentration changes become extreme?
Answer: The reaction can become less favorable and may eventually stop or reverse as conditions depart far from standard This question assesses the impact of concentration changes on cell potential and reaction spontaneity in galvanic cells. The Nernst equation shows that cell potential depends on the reaction quotient Q, where deviations from standard concentrations can alter the driving force of the redox reaction. In this Zn-Cu galvanic cell, initial standard conditions favor Zn oxidation and Cu²⁺ reduction, but as Zn²⁺ concentration increases and Cu²⁺ decreases, Q rises, reducing the cell potential. Choice A is correct because extreme concentration changes can make the reaction less favorable, potentially reaching equilibrium (E=0) or even reversing if Q becomes sufficiently large to make E negative. Choice B fails by assuming anode and cathode roles are fixed regardless of conditions, ignoring how concentrations influence reaction direction via the Nernst equation. For similar scenarios, calculate the cell potential using the Nernst equation to predict spontaneity. Always consider that when Q > K, the reverse reaction becomes favored, providing a check for potential reversal in electrochemical systems.
A lab models mitochondrial electron transfer using a galvanic cell with half-cells: Co $^{3+}$ /Co $^{2+}$ and Fe $^{3+}$ /Fe $^{2+}$ , each at 1.0 M. Standard reduction potentials: Co $^{3+}$ /Co $^{2+}$ = $+1.82$ V; Fe $^{3+}$ /Fe $^{2+}$ = $+0.77$ V. Which statement best reflects the redox principle illustrated?
Answer: Fe $^{2+}$ is oxidized and Co $^{3+}$ is reduced; electrons flow from Fe $^{2+}$ to Co $^{3+}$ The skill being tested is determining spontaneous electron transfer in a galvanic cell with ion-only half-cells using standard reduction potentials. In such cells, the couple with the higher E° acts as the oxidizing agent (reduced), while the lower E° couple is oxidized, with electrons flowing from the reducing agent to the oxidizing agent. Here, Co³⁺/Co²⁺ has E° = +1.82 V (higher than Fe³⁺/Fe²⁺'s +0.77 V), so Fe²⁺ is oxidized to Fe³⁺, and Co³⁺ is reduced to Co²⁺, with electrons flowing from Fe to Co. This makes choice A correct, identifying Fe²⁺ oxidation and Co³⁺ reduction with correct electron flow. Choice B fails because it incorrectly states Co²⁺ is reduced to Co³⁺, but the higher E° means Co³⁺ is reduced. In similar redox scenarios, identify the stronger oxidizing agent as the one with higher E° (cathode). Check for spontaneity by ensuring E°_cell = E°_cathode - E°_anode > 0.
A galvanic cell uses the half-cells: Cd(s)|Cd $^{2+}$ and Cu(s)|Cu $^{2+}$ , each 1.0 M. Standard reduction potentials: Cd $^{2+}$ /Cd = $-0.40$ V; Cu $^{2+}$ /Cu = $+0.34$ V. Which conclusion is most consistent with the electrochemical process at standard conditions?
Answer: Cd(s) is oxidized at the anode, and the Cd electrode mass decreases The skill being tested is determining electrode mass changes and spontaneity in a galvanic cell from standard reduction potentials. In galvanic cells, the anode metal loses mass due to oxidation if its E° is lower, and the cell is spontaneous if E°_cell > 0. Here, Cd has E° = -0.40 V (lower than Cu's +0.34 V), so Cd(s) is oxidized at the anode, decreasing its mass, and the cell runs spontaneously. This makes choice C correct, identifying Cd oxidation and mass decrease. Choice B fails because Cu has higher E°, so it's reduced at the cathode, not oxidized. For similar redox scenarios, identify mass loss at the anode where oxidation dissolves the metal. Confirm spontaneity with E°_cell > 0; external voltage is only for E°_cell < 0.
In a benchtop study of oxidative stress, a galvanic cell is assembled to model electron transfer from a metal surface to an oxidant. The half-cells are: Zn(s)|Zn $^{2+}$ (1.0 M) and Ag $^+$ (1.0 M)|Ag(s), connected by a salt bridge at 25°C. Standard reduction potentials: $E^\circ$ (Ag $^+$ /Ag) = +0.80 V; $E^\circ$ (Zn $^{2+}$ /Zn) = −0.76 V. Which outcome would be expected in this electrochemical cell under standard conditions?
Answer: Electrons flow from Zn to Ag, and the Zn electrode loses mass as Zn(s) is oxidized This question tests understanding of spontaneous electron flow in galvanic cells based on reduction potentials. In a galvanic cell, electrons flow spontaneously from the species with the lower (more negative) reduction potential to the species with the higher (more positive) reduction potential. Since Zn²⁺/Zn has E° = -0.76 V and Ag⁺/Ag has E° = +0.80 V, zinc will be oxidized at the anode (Zn → Zn²⁺ + 2e⁻) and silver ions will be reduced at the cathode (Ag⁺ + e⁻ → Ag). This means electrons flow from Zn to Ag through the external circuit, and the Zn electrode loses mass as solid zinc is oxidized to Zn²⁺ ions that enter solution. Choice C incorrectly identifies Ag as being oxidized when it is actually being reduced and plated out. To identify electron flow direction in any galvanic cell, compare reduction potentials: electrons always flow from the more negative to the more positive E° value.
A microelectrode experiment uses a galvanic cell with H $_2$ (g, 1 atm)|H $^+$ (1.0 M) and Fe $^{3+}$ (1.0 M)/Fe $^{2+}$ (1.0 M) coupled via inert Pt electrodes. Standard reduction potentials: 2H $^+$ + 2e $^-$ $\rightarrow$ H $_2$ , $E^\circ=0.00$ V; Fe $^{3+}$ + e $^-$ $\rightarrow$ Fe $^{2+}$ , $E^\circ=+0.77$ V. Which outcome would be expected?
Answer: H $_2$ is oxidized at the anode, and Fe $^{3+}$ is reduced at the cathode The skill being tested is identifying redox processes involving the standard hydrogen electrode in a galvanic cell. In galvanic cells with SHE (E° = 0.00 V), if paired with a higher E° couple, SHE acts as anode (H₂ oxidized); the other is cathode. Here, Fe³⁺/Fe²⁺ has E° = +0.77 V > 0.00 V, so H₂ is oxidized at the anode, and Fe³⁺ is reduced at the cathode. This makes choice A correct, stating H₂ oxidation at anode and Fe³⁺ reduction at cathode. Choice D fails because electrons flow from anode (SHE) to cathode (Fe), not vice versa. In similar redox scenarios, use SHE as reference: positive E° means reduction at that electrode. Check direction by electron flow from lower E° to higher E° externally.
An electrochemical sensor uses a galvanic cell with half-cells: Al(s)|Al $^{3+}$ (1.0 M) and Ag $^+$ (1.0 M)|Ag(s) at 25°C. Standard reduction potentials: $E^\circ$ (Ag $^+$ /Ag) = +0.80 V; $E^\circ$ (Al $^{3+}$ /Al) = −1.66 V. Which statement best reflects the redox principle illustrated?
Answer: Ag $^+$ is reduced at the cathode because it has the more positive reduction potential This question tests understanding of cathode identification based on reduction potentials. In a galvanic cell, reduction always occurs at the cathode, and the species with the more positive reduction potential undergoes reduction. Since Ag⁺/Ag has E° = +0.80 V (positive) and Al³⁺/Al has E° = -1.66 V (negative), Ag⁺ will be reduced to Ag at the cathode while Al will be oxidized to Al³⁺ at the anode. The large potential difference (2.46 V) drives vigorous electron flow from Al to Ag through the external circuit. Choice C incorrectly states that electrons flow from Ag to Al and misidentifies the cathode as the electron source; in reality, the anode (Al) is the source of electrons. To identify the cathode in any galvanic cell, look for the half-reaction with the more positive (or less negative) reduction potential—this species gains electrons and is reduced.
A researcher builds a galvanic cell to power a low-current microfluidic pump: Sn(s)|Sn $^{2+}$ (1.0 M) || Pb $^{2+}$ (1.0 M)|Pb(s). Standard reduction potentials: Sn $^{2+}$ /Sn = $-0.14$ V; Pb $^{2+}$ /Pb = $-0.13$ V. Based on the setup, which conclusion is most consistent with the electrochemical process at standard conditions?
Answer: Sn(s) is oxidized and electrons flow from Sn to Pb, giving a small positive $E^\circ_{\text{cell}}$ The skill being tested is predicting the spontaneous redox reaction and electron flow in a galvanic cell with closely valued reduction potentials. In galvanic cells, even small differences in E° determine direction: the half-cell with slightly higher E° is the cathode (reduction), and the lower is the anode (oxidation), with electrons flowing from anode to cathode. Here, Pb has E° = -0.13 V (slightly higher than Sn's -0.14 V), so Sn(s) is oxidized at the anode, Pb²⁺ is reduced at the cathode, and E°_cell is small but positive. This makes choice A correct, as Sn(s) oxidation and electron flow from Sn to Pb yield a small positive E°_cell. Choice C fails because it reverses electron flow; electrons flow from lower E° (Sn) to higher (Pb). In similar redox scenarios, subtract E°_anode from E°_cathode to get E°_cell and confirm positivity for spontaneity. For close values, note that the more negative E° indicates easier oxidation.
A galvanic cell is constructed as Al(s)|Al $^{3+}$ (1.0 M)||Ni $^{2+}$ (1.0 M)|Ni(s) at 25°C. Standard reduction potentials: Al $^{3+}$ /Al = −1.66 V; Ni $^{2+}$ /Ni = −0.25 V. Based on the setup, which outcome would be expected in this electrochemical cell under standard conditions?
Answer: Al(s) is oxidized at the anode, and electrons flow through the wire from Al to Ni. This question tests understanding of galvanic cells with negative reduction potentials. Even when both reduction potentials are negative, the more negative value indicates the stronger reducing agent that will be oxidized at the anode. Since Al³⁺/Al (E° = -1.66 V) is more negative than Ni²⁺/Ni (E° = -0.25 V), Al(s) is oxidized to Al³⁺ at the anode while Ni²⁺ is reduced to Ni(s) at the cathode. Electrons flow from the Al anode through the external wire to the Ni cathode, and E°cell = -0.25 - (-1.66) = +1.41 V, confirming spontaneous operation. Choice A incorrectly identifies Ni as the anode, choice C incorrectly identifies Al as the cathode, and choice D incorrectly concludes the cell is nonspontaneous (the cell is spontaneous because E°cell > 0). For cells with negative E° values: more negative = anode (oxidation), less negative = cathode (reduction), and E°cell can still be positive.
A student reverses the connections on a working galvanic cell (swaps the voltmeter leads) without changing the chemistry. Which conclusion is most consistent with the electrochemical process and the measured voltage?
Answer: The sign of the measured cell potential reverses, but the direction of spontaneous electron flow in the cell does not This question tests understanding of voltmeter polarity and electron flow in galvanic cells. When voltmeter leads are reversed on a functioning galvanic cell, the measured voltage changes sign (positive becomes negative or vice versa) but the actual electron flow in the cell remains unchanged. The chemical reactions at the anode and cathode continue in the same direction because they are determined by the reduction potentials, not by how the measuring device is connected. This makes option A correct: the sign of the measured potential reverses, but spontaneous electron flow continues in the same direction. Option B incorrectly suggests the cell becomes electrolytic, while option C wrongly claims electrodes swap roles. Remember: measurement setup doesn't change cell chemistry; it only affects how we read the voltage.
In a conceptual prediction, a galvanic cell uses Zn(s)|Zn $^{2+}$ and Cu(s)|Cu $^{2+}$ . The salt bridge contains KNO $_3$ . Which outcome would be expected as the cell operates to maintain electroneutrality?
Answer: NO $_3^-$ migrates toward the anode compartment where Zn $^{2+}$ accumulates The skill being tested is understanding ion migration in a salt bridge to maintain electroneutrality in a galvanic cell. In galvanic cells, anions migrate toward the anode to balance positive ion buildup from oxidation, while cations migrate toward the cathode to balance negative charge from reduction. In this Zn/Cu cell, Zn²⁺ accumulates at the anode, so NO₃⁻ migrates toward the anode compartment. This makes choice B correct, as NO₃⁻ moves to the anode where Zn²⁺ accumulates. Choice C fails because NO₃⁻ migrates to the anode, not cathode; cathode needs cations to balance electron influx. For similar redox scenarios, identify ion flow: anions to anode, cations to cathode. Check electroneutrality by noting oxidation produces cations at anode, reduction consumes cations at cathode.
A researcher couples two half-cells to mimic cellular redox buffering: Fe $^{3+}$ /Fe $^{2+}$ (1.0 M each) and Cu $^{2+}$ /Cu(s) (1.0 M). Standard reduction potentials: Fe $^{3+}$ /Fe $^{2+}$ = +0.77 V; Cu $^{2+}$ /Cu = +0.34 V. Based on the setup, which outcome would be expected in this galvanic cell under standard conditions?
Answer: Fe $^{3+}$ is reduced to Fe $^{2+}$ at the cathode, and Cu(s) is oxidized to Cu $^{2+}$ at the anode. This question tests identification of redox reactions in a galvanic cell with two positive reduction potentials. In galvanic cells, the half-reaction with the higher (more positive) reduction potential occurs as reduction at the cathode, while the half-reaction with the lower (less positive) reduction potential must be reversed to oxidation at the anode. Since Fe³⁺/Fe²⁺ has E° = +0.77 V (higher) and Cu²⁺/Cu has E° = +0.34 V (lower), Fe³⁺ is reduced to Fe²⁺ at the cathode while Cu(s) is oxidized to Cu²⁺ at the anode. Choice B incorrectly assigns the higher potential species to oxidation, choice C confuses electron flow direction (electrons flow from lower to higher E°), and choice D incorrectly claims both undergo reduction in a galvanic cell. When both E° values are positive, compare their magnitudes: higher E° = cathode (reduction), lower E° = anode (oxidation reversed), ensuring spontaneous operation with positive E°cell.
In a respiration-inspired assay, a mediator M(ox)/M(red) has $E^{\circ\prime} = +0.05$ V (as a reduction). It is mixed with FAD/FADH $_2$ with $E^{\circ\prime} = -0.22$ V. Which statement best reflects the redox principle illustrated?
Answer: FADH $_2$ will be oxidized and will reduce M(ox) to M(red) This question tests understanding of electron flow direction in biological redox systems. Spontaneous electron transfer occurs from the species with the more negative reduction potential to the species with the more positive reduction potential. Since E°'(FAD/FADH₂) = -0.22 V < E°'(M(ox)/M(red)) = +0.05 V, FADH₂ will spontaneously donate electrons (be oxidized to FAD) while M(ox) accepts electrons (is reduced to M(red)). This makes option A correct, with FADH₂ acting as the electron donor and M(ox) as the electron acceptor. Option B incorrectly reverses the electron flow based on the potentials given. For biological electron transport, remember: electrons flow from more negative E°' to more positive E°'.
A clinician-scientist evaluates a point-of-care galvanic sensor whose output depends on ion concentrations. The cell is based on Ag $^+$ /Ag and is initially assembled under standard conditions. During testing, $[\text{Ag}^+]$ in the cathode compartment is decreased (all else constant, $T=25^\circ\text{C}$ ). Using $E=E^\circ-\frac{0.059}{n}\log Q$ , which outcome would be expected in this electrochemical cell?
Answer: The cell voltage increases because lowering product concentration drives reduction and makes $E$ more positive. This question tests understanding of the Nernst equation and how concentration changes affect cell voltage. The Nernst equation E = E° - (0.059/n)log Q shows that decreasing [Ag⁺] at the cathode decreases Q (since [Ag⁺] appears in the denominator of Q for the reduction reaction). When Q decreases, log Q becomes more negative, making the term -(0.059/n)log Q more positive, thus increasing E. This makes physical sense: lowering product concentration drives the reduction reaction forward, making the cell voltage more positive. The correct answer A accurately describes this: the cell voltage increases because lowering product concentration drives reduction and makes E more positive. Answer B incorrectly states that lowering [Ag⁺] increases Q - for the reduction reaction Ag⁺ + e⁻ → Ag, Q = 1/[Ag⁺], so decreasing [Ag⁺] decreases Q. When analyzing concentration effects, remember that decreasing products or increasing reactants at the cathode increases cell voltage.
An electrolytic cell is used to regenerate NADH analogs by forcing reduction of a substrate at the cathode. Which statement best reflects the redox principle illustrated in electrolytic cells?
Answer: Reduction occurs at the cathode even in electrolytic cells, but the overall process is nonspontaneous The skill being tested is distinguishing redox processes and spontaneity in electrolytic cells. In electrolytic cells, reduction still occurs at the cathode and oxidation at the anode, but the overall reaction is nonspontaneous, requiring external voltage. In this NADH regeneration setup, forced reduction at the cathode exemplifies nonspontaneous electrolysis with standard electrode definitions. This makes choice B correct, as reduction is at the cathode, but the process is nonspontaneous. Choice A fails because oxidation is at the anode, not cathode, regardless of terminal. For similar redox scenarios, remember cathode = reduction in all cells. Check for electrolysis if E°_cell < 0, needing external energy.
A researcher models mitochondrial electron transport using a simplified redox pair: NADH/NAD $^+$ and O $_2$ /H $_2$ O at pH 7. Standard reduction potentials (biochemical standard state): $E^{\circ\prime}$ (NAD $^+$ /NADH) = −0.32 V; $E^{\circ\prime}$ (O $_2$ /H $_2$ O) = +0.82 V. Which statement best reflects the redox principle illustrated by these values?
Answer: Electrons flow spontaneously from NADH to O $_2$ because O $_2$ /H $_2$ O has the higher $E^{\circ\prime}$ This question tests understanding of electron flow direction based on biochemical reduction potentials. In any redox system, electrons flow spontaneously from the half-reaction with the lower (more negative) reduction potential to the half-reaction with the higher (more positive) reduction potential. Since NAD⁺/NADH has E°' = -0.32 V and O₂/H₂O has E°' = +0.82 V, electrons will flow from NADH (which gets oxidized to NAD⁺) to O₂ (which gets reduced to H₂O). This mirrors the actual electron transport chain in mitochondria where NADH serves as an electron donor and oxygen as the terminal electron acceptor. Choice A incorrectly identifies NADH as an oxidizing agent when it is actually a reducing agent (electron donor). To predict electron flow in biological systems, identify the species with the more negative E°' as the electron donor (reducing agent) and the species with the more positive E°' as the electron acceptor (oxidizing agent).
A researcher observes that in a galvanic cell, the mass of one metal electrode increases over time while the other decreases. Which conclusion is most consistent with the electrochemical process causing the mass increase?
Answer: The electrode gaining mass is the cathode where metal ions are reduced and plate as solid This question tests understanding of mass changes at electrodes in galvanic cells. In galvanic cells, the cathode is where reduction occurs: metal cations in solution gain electrons and deposit as solid metal atoms on the electrode surface, increasing its mass. Conversely, at the anode, solid metal atoms lose electrons and enter solution as cations, decreasing electrode mass. This makes option B correct, identifying the mass-gaining electrode as the cathode where metal ion reduction and plating occur. Option A incorrectly identifies this as the anode, while option C confuses galvanic cells with electrolytic cells. For electrode mass changes: cathode gains mass (reduction/plating), anode loses mass (oxidation/dissolution).
In an electrolytic cell used to generate H $_2$ from water at near-neutral pH, a cathode catalyzes 2H $_2$ O + 2e $^-$ $\rightarrow$ H $_2$ + 2OH $^-$ . Which conclusion is most consistent with the electrochemical process at the cathode?
Answer: Reduction occurs at the cathode, consuming electrons supplied by the power source This question tests understanding of cathode processes in electrolytic cells. In any electrochemical cell (galvanic or electrolytic), reduction always occurs at the cathode by definition. In this water electrolysis cell, the cathode reaction 2H₂O + 2e⁻ → H₂ + 2OH⁻ shows water molecules gaining electrons (reduction) to produce hydrogen gas. The external power source supplies these electrons to the cathode, forcing this nonspontaneous reduction. This makes option B correct, stating that reduction occurs at the cathode with electrons supplied by the power source. Option A incorrectly claims oxidation at the cathode, while option C misidentifies the reactant. Remember the fundamental rule: cathode = reduction (gain of electrons), regardless of cell type.
A student investigates corrosion protection by constructing a galvanic cell with Mg(s)|Mg $^{2+}$ (1.0 M) and Ni(s)|Ni $^{2+}$ (1.0 M). Standard reduction potentials: Mg $^{2+}$ /Mg = $-2.37$ V; Ni $^{2+}$ /Ni = $-0.25$ V. Which statement best reflects the redox principle illustrated?
Answer: Mg(s) is oxidized, serving as the anode, and protects Ni by sacrificing electrons The skill being tested is applying redox principles to cathodic protection in a galvanic cell setup using standard reduction potentials. In galvanic cells for corrosion protection, the metal with the more negative E° acts as the sacrificial anode, oxidizing preferentially to protect the other metal. In this Mg/Ni cell, Mg has E° = -2.37 V (more negative than Ni's -0.25 V), so Mg(s) is oxidized at the anode, protecting Ni by donating electrons. This makes choice B correct, as Mg serves as the anode and sacrifices electrons to prevent Ni oxidation. Choice A fails because it incorrectly identifies Ni as the anode due to its more positive E°, but the anode is actually the one more easily oxidized (lower E°). In similar redox scenarios, identify the sacrificial anode as the metal with the lower E°. Check that electron flow is from the anode (lower E°) to the cathode (higher E°) to confirm protective current direction.
In a benchtop study of a glucose biosensor, a galvanic cell is assembled with a Zn(s)|Zn $^{2+}$ (1.0 M) half-cell and a Cu(s)|Cu $^{2+}$ (1.0 M) half-cell connected by a salt bridge. Standard reduction potentials are: Zn $^{2+}$ /Zn = $-0.76$ V and Cu $^{2+}$ /Cu = $+0.34$ V. Based on the setup, which outcome would be expected in this electrochemical cell at standard conditions?
Answer: Electrons flow through the wire from Zn to Cu, and the Zn electrode loses mass The skill being tested is identifying the direction of electron flow and mass changes in a galvanic cell based on standard reduction potentials. In galvanic cells, the half-cell with the lower (more negative) reduction potential acts as the anode where oxidation occurs, while the higher potential half-cell is the cathode where reduction happens, with electrons flowing from anode to cathode. In this Zn/Cu cell, Zn has E° = -0.76 V (lower than Cu's +0.34 V), so Zn is oxidized at the anode and Cu²⁺ is reduced at the cathode. Thus, electrons flow through the wire from Zn to Cu, and the Zn electrode loses mass as Zn(s) becomes Zn²⁺(aq), making choice B correct. Choice A fails because it reverses the oxidation and reduction sites, as Zn is oxidized, not reduced. To check similar redox scenarios, always identify the anode as the half-cell with the lower E° and confirm electron flow from anode to cathode. Additionally, calculate E°_cell = E°_cathode - E°_anode to ensure it's positive for spontaneity.
A galvanic cell is assembled with half-cells: Pb(s)|Pb $^{2+}$ and Ag(s)|Ag $^{+}$ (both 1.0 M). Standard reduction potentials: Pb $^{2+}$ /Pb = $-0.13$ V; Ag $^{+}$ /Ag = $+0.80$ V. Which statement best reflects the redox principle illustrated?
Answer: Ag $^{+}$ is reduced at the cathode; Pb(s) is oxidized at the anode The skill being tested is applying standard reduction potentials to identify cathode and anode in a galvanic cell. In galvanic cells, the half-cell with the higher E° is the cathode (reduction), and the lower E° is the anode (oxidation). Here, Ag has E° = +0.80 V (higher than Pb's -0.13 V), so Ag⁺ is reduced at the cathode, and Pb(s) is oxidized at the anode. This makes choice B correct, identifying Ag⁺ reduction at cathode and Pb oxidation at anode. Choice A fails because Pb²⁺ has a more negative E°, making it less likely to be reduced; it's oxidized instead. For similar redox scenarios, compare E° to assign roles: higher E° = cathode. Verify by calculating E°_cell = E°_cathode - E°_anode > 0 for spontaneity.

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