MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4c Electrochemical Cells Redox

What this quiz covers

This quiz focuses on 4c Electrochemical Cells Redox, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Electrons flow from the 1.0 M side to the 0.10 M side because higher ion concentration favors oxidation.
2. Electrons flow from the 0.10 M side to the 1.0 M side, increasing Cu$^{2+}$ on the dilute side.
3. Electrons flow from the 0.10 M side to the 1.0 M side, decreasing Cu$^{2+}$ on the concentrated side. (correct answer)
4. No voltage is produced because both half-cells use the same Cu$^{2+}$/Cu redox couple.

Explanation: This question tests understanding of concentration cells where identical electrodes create voltage from concentration differences alone. In a concentration cell, the dilute side (0.10 M Cu²⁺) acts as the anode where Cu(s) → Cu²⁺ + 2e⁻ occurs to increase [Cu²⁺], while the concentrated side (1.0 M Cu²⁺) acts as the cathode where Cu²⁺ + 2e⁻ → Cu(s) occurs to decrease [Cu²⁺]. Electrons flow from the anode (0.10 M side) to the cathode (1.0 M side) through the external wire, and this process continues until concentrations equalize. The net effect is that Cu²⁺ concentration increases on the dilute side and decreases on the concentrated side. Choice A reverses the electron flow direction, choice B correctly identifies flow direction but incorrectly states the concentration change, and choice D incorrectly claims no voltage (concentration differences create voltage even with identical E° values). For concentration cells: electrons flow from dilute to concentrated, equalizing concentrations.

Question 2

1. Electrons flow from the Fe$^{3+}$/Fe$^{2+}$half-cell to the Cu$^{2+}$/Cu half-cell because Fe$^{3+}$ is the stronger oxidizing agent.
2. Cu(s) is oxidized at the anode, and Fe$^{3+}$ is reduced at the cathode in the spontaneous cell. (correct answer)
3. Fe$^{2+}$ is oxidized at the anode, and Cu$^{2+}$ is reduced at the cathode in the spontaneous cell.
4. The cell is nonspontaneous because both listed $E^\circ$ values are positive, so $E^\circ_{\text{cell}}$ must be negative.

Explanation: This question tests the ability to identify spontaneous electron flow in a galvanic cell based on standard reduction potentials. In a galvanic cell, the half-reaction with the higher (more positive) reduction potential occurs as reduction at the cathode, while the half-reaction with the lower reduction potential runs in reverse as 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. The correct answer B accurately describes this: Cu(s) is oxidized at the anode and Fe³⁺ is reduced at the cathode. Answer A incorrectly states electron flow direction - electrons actually flow from the Cu/Cu²⁺ half-cell (anode) to the Fe³⁺/Fe²⁺ half-cell (cathode). To identify the spontaneous direction in any galvanic cell, compare E° values: the higher E° species undergoes reduction at the cathode, while the lower E° species undergoes oxidation at the anode.

Question 3

1. The dilute side is the cathode because reduction is favored at lower cation concentration.
2. The concentrated side is the cathode, and Cu$^{2+}$ is reduced there, increasing the Cu$^{2+}$ gradient over time.
3. The concentrated side is the cathode, and Cu$^{2+}$ is reduced there, decreasing the concentration difference over time. (correct answer)
4. No voltage is produced because identical metals are used, so $E=0$ regardless of concentration.

Explanation: This question tests understanding of concentration cells, where identical electrodes in different ion concentrations create a voltage. In a concentration cell, the Nernst equation shows that the half-cell with higher ion concentration acts as the cathode (reduction occurs), while the half-cell with lower ion concentration acts as the anode (oxidation occurs). For this Cu/Cu²⁺ concentration cell, the 1.0 M side is the cathode where Cu²⁺ + 2e⁻ → Cu(s) occurs, while the 0.010 M side is the anode where Cu(s) → Cu²⁺ + 2e⁻ occurs. This process naturally decreases the concentration difference over time as Cu²⁺ is consumed at the cathode and produced at the anode. The correct answer C accurately describes this: the concentrated side is the cathode where Cu²⁺ is reduced, decreasing the concentration difference. Answer B incorrectly states that reduction would increase the gradient - reduction consumes Cu²⁺, decreasing its concentration. In any concentration cell, the driving force is to equalize concentrations, with reduction occurring at the higher concentration side.

Question 4

1. Ni(s) dissolves at the implant because the implant is the anode in electroplating.
2. Ni$^{2+}$ is reduced to Ni(s) on the implant surface, requiring electrons supplied to the cathode by the power source. (correct answer)
3. Ni$^{2+}$ is oxidized to Ni$^{3+}$ at the implant because reduction occurs at the anode in an electrolytic cell.
4. Ni$^{2+}$ is reduced at the Pt anode because inert electrodes are always the site of reduction.

Explanation: This question tests understanding of electrolytic cells used for electroplating, where an external voltage drives a nonspontaneous redox reaction. In electroplating, the object to be plated (the implant) serves as the cathode where reduction occurs, while oxidation occurs at the anode. Since Ni²⁺ + 2e⁻ → Ni(s) has E° = -0.25 V (negative), this reduction is nonspontaneous and requires an external power source to supply electrons to the cathode. At the cathode (implant), Ni²⁺ ions gain electrons and are reduced to metallic Ni(s), which deposits on the implant surface. The correct answer B accurately describes this process: Ni²⁺ is reduced to Ni(s) on the implant surface with electrons supplied by the power source. Answer A incorrectly suggests Ni dissolves at the implant - dissolution would occur if the implant were the anode, not the cathode. In any electroplating setup, remember that the cathode is where metal deposition occurs through reduction, requiring an external voltage when E° is negative.

Question 5

1. Oxidation occurs at the cathode, so Zn(s) gains electrons and Ag$^+$ loses electrons in the spontaneous cell.
2. Electrons flow through the wire from Ag to Zn because Ag has the higher $E^\circ$ value.
3. Zn(s) is oxidized at the anode, providing electrons that reduce Ag$^+$ at the cathode. (correct answer)
4. The salt bridge supplies electrons to complete the circuit, so no ion migration is needed.

Explanation: This question tests the fundamental principle of electron flow in galvanic cells based on reduction potentials. In a spontaneous galvanic cell, the species with the more negative reduction potential is oxidized at the anode, while the species with the more positive reduction potential is reduced at the cathode. Since Zn²⁺/Zn has E° = -0.76 V (more negative) and Ag⁺/Ag has E° = +0.80 V (more positive), Zn(s) spontaneously loses electrons (oxidation) at the anode while Ag⁺ gains electrons (reduction) at the cathode. The electrons flow through the external wire from the Zn electrode (anode) to the Ag electrode (cathode). The correct answer C accurately describes this: Zn(s) is oxidized at the anode, providing electrons that reduce Ag⁺ at the cathode. Answer B incorrectly states the electron flow direction - electrons flow from Zn to Ag, not from Ag to Zn. To predict electron flow in any galvanic cell, identify the more negative E° species as the anode (source of electrons) and the more positive E° species as the cathode (sink for electrons).

Question 6

1. O$_2$ is oxidized at the anode because it has the larger positive $E^\circ$ value.
2. Electrons flow from the O$_2$/H$_2$O half-cell to the mediator half-cell because oxygen is the strongest oxidant present.
3. The mediator is reduced at the cathode and O$_2$ is reduced at the anode because both half-reactions proceed as reductions.
4. The mediator (Red form) is oxidized at the anode, supplying electrons that reduce O$_2$ at the cathode. (correct answer)

Explanation: This question tests understanding of redox reactions in biological assay systems using mediator couples. In this galvanic cell, O₂/H₂O has E° = +1.23 V (much higher) compared to the mediator Ox/Red couple with E° = +0.10 V (lower), establishing O₂ as the stronger oxidizing agent. At the cathode, O₂ undergoes reduction: O₂ + 4H⁺ + 4e⁻ → 2H₂O, while at the anode, the reduced form of the mediator undergoes oxidation: Red → Ox + e⁻. Electrons flow from the mediator half-cell (anode) to the O₂/H₂O half-cell (cathode). The correct answer D accurately describes this: the mediator (Red form) is oxidized at the anode, supplying electrons that reduce O₂ at the cathode. Answer A incorrectly suggests O₂ is oxidized - with the highest E° value, O₂ is the strongest oxidant and must be reduced, not oxidized. In biological redox assays, mediators with intermediate E° values facilitate electron transfer between enzymes and electrodes.

Question 7

1. Mg is the cathode because it has the more negative $E^\circ$, so it more strongly attracts electrons.
2. Cu is oxidized at the anode because it has the more positive $E^\circ$ value.
3. Electrons flow from Mg to Cu through the external circuit, and Mg(s) mass decreases during operation. (correct answer)
4. The salt bridge provides electrons to Cu$^{2+}$, so no metal oxidation is required.

Explanation: This question tests understanding of electron flow and mass changes in galvanic cells involving solid metal electrodes. In this Mg/Cu galvanic cell, Mg²⁺/Mg has E° = -2.37 V (very negative) while Cu²⁺/Cu has E° = +0.34 V (positive), creating a large driving force for spontaneous reaction. The more negative Mg acts as the anode where Mg(s) → Mg²⁺ + 2e⁻ (oxidation), causing the Mg electrode to lose mass as metal atoms enter solution. Electrons flow through the external circuit from Mg (anode) to Cu (cathode), where Cu²⁺ + 2e⁻ → Cu(s) occurs. The correct answer C accurately describes this: electrons flow from Mg to Cu through the external circuit, and Mg(s) mass decreases during operation. Answer A incorrectly identifies Mg as the cathode - the more negative reduction potential means Mg is more easily oxidized, making it the anode. In any galvanic cell with solid metal electrodes, the anode loses mass through oxidation while the cathode gains mass through reduction.

Question 8

1. Sn$^{4+}$ is reduced at the cathode because it requires more electrons per ion than Fe$^{3+}$.
2. Fe$^{2+}$ is oxidized to Fe$^{3+}$ at the anode, while Sn$^{4+}$ is reduced to Sn$^{2+}$ at the cathode.
3. Sn$^{2+}$ is oxidized to Sn$^{4+}$ at the anode, while Fe$^{3+}$ is reduced to Fe$^{2+}$ at the cathode. (correct answer)
4. Both Sn$^{4+}$ and Fe$^{3+}$ are reduced at their respective electrodes because both $E^\circ$ values are positive.

Explanation: This question tests identification of redox reactions when both reduction potentials are positive. Comparing the standard reduction potentials, Fe³⁺/Fe²⁺ (E° = +0.77 V) is more positive than Sn⁴⁺/Sn²⁺ (E° = +0.15 V), so Fe³⁺ undergoes reduction at the cathode while the Sn⁴⁺/Sn²⁺ half-reaction must be reversed to oxidation at the anode. At the cathode: Fe³⁺ + e⁻ → Fe²⁺ (reduction), and at the anode: Sn²⁺ → Sn⁴⁺ + 2e⁻ (oxidation). The spontaneous cell reaction is 2Fe³⁺ + Sn²⁺ → 2Fe²⁺ + Sn⁴⁺ with E°cell = 0.77 - 0.15 = 0.62 V. Choice A incorrectly assigns reduction based on electron stoichiometry rather than E° values, choice B reverses the correct assignments, and choice D incorrectly claims both undergo reduction. When comparing positive E° values: higher E° = reduction at cathode, lower E° = oxidation at anode (half-reaction reversed).

Question 9

1. The electrode gaining mass is the anode where metal atoms oxidize into solution
2. The electrode gaining mass is the cathode where metal ions are reduced and plate as solid (correct answer)
3. The electrode gaining mass must be connected to the negative terminal of a power supply
4. Mass increase indicates electrons are being produced there by reduction

Explanation: 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).

Question 10

1. Br$^-$ is reduced to Br$_2$ at the cathode and Co(s) is reduced at the anode
2. Co(s) is oxidized at the anode and Br$_2$ is reduced to Br$^-$ at the cathode (correct answer)
3. Electrons flow from bromine to cobalt because halogens are strong reducing agents
4. Co$^{2+}$ is reduced at the cathode because it is the only cation present

Explanation: This question tests understanding of galvanic cells involving halogens. The half-reaction with the more positive reduction potential proceeds as reduction at the cathode. Since E°(Br₂/Br⁻) = +1.07 V > E°(Co²⁺/Co) = -0.28 V, Br₂ will be reduced to Br⁻ at the cathode while Co(s) will be oxidized to Co²⁺ at the anode. This makes option B correct, with Co(s) oxidation at the anode and Br₂ reduction at the cathode, with electrons flowing from Co to the Pt electrode. Option A incorrectly suggests Br⁻ oxidation when Br₂ is already present. For halogen systems, remember that X₂ + 2e⁻ → 2X⁻ is the reduction; the reverse is oxidation.

Question 11

1. Fe$^{3+}$ is the oxidizing agent because it is reduced during the reaction (correct answer)
2. Ascorbate is the oxidizing agent because it loses electrons
3. Fe$^{2+}$ is the oxidizing agent because it has fewer positive charges
4. Both reactants are oxidizing agents because electron transfer is bidirectional

Explanation: This question tests understanding of oxidizing and reducing agent terminology in redox reactions. In any redox reaction, the species that gains electrons (undergoes reduction) is the oxidizing agent, while the species that loses electrons (undergoes oxidation) is the reducing agent. Since Fe³⁺ is reduced to Fe²⁺ (gains an electron), Fe³⁺ acts as the oxidizing agent. Conversely, ascorbate loses electrons (is oxidized) and thus acts as the reducing agent. This makes option A correct, identifying Fe³⁺ as the oxidizing agent because it undergoes reduction. Option B incorrectly labels ascorbate as the oxidizing agent despite it being oxidized. To identify agents: oxidizing agents get reduced (gain electrons), reducing agents get oxidized (lose electrons).

Question 12

1. Fe(s) is reduced at the cathode, causing Fe plating, because it has the more negative $E^\circ$ value.
2. Cu$^{2+}$ is reduced at the cathode, and Fe(s) is oxidized at the anode, consistent with spontaneous corrosion of iron. (correct answer)
3. Electrons flow from Cu to Fe because Cu$^{2+}$ has a higher tendency to be oxidized.
4. The anode is the site of reduction in a galvanic cell, so Cu$^{2+}$ must be reduced at the anode.

Explanation: This question tests understanding of spontaneous corrosion processes modeled by galvanic cells. In this Fe/Cu cell, Fe²⁺/Fe has E° = -0.44 V (more negative) while Cu²⁺/Cu has E° = +0.34 V (more positive), making iron the anode where oxidation occurs. At the anode, Fe(s) → Fe²⁺ + 2e⁻ (iron corrosion/oxidation), while at the cathode, Cu²⁺ + 2e⁻ → Cu(s) (copper ion reduction). This models the spontaneous corrosion of iron in the presence of a more noble metal like copper. The correct answer B accurately describes this: Cu²⁺ is reduced at the cathode and Fe(s) is oxidized at the anode, consistent with spontaneous iron corrosion. Answer D incorrectly states that reduction occurs at the anode - in galvanic cells, oxidation always occurs at the anode and reduction at the cathode. This principle explains why iron corrodes preferentially when in contact with more noble metals in marine environments.

Question 13

1. O$_2$ is reduced (oxidant) and GSH is oxidized (reductant) (correct answer)
2. O$_2$ is oxidized because it has the larger positive $E^\circ$
3. GSSG is oxidized to GSH at the anode because anodes attract anions
4. Electron flow is impossible because one half-reaction involves protons

Explanation: This question tests understanding of redox roles in biological oxidation reactions. In spontaneous redox reactions, the species with the more positive reduction potential acts as the oxidizing agent (gets reduced), while the species with the more negative potential acts as the reducing agent (gets oxidized). Since E°(O₂/H₂O) = +1.23 V >> E°(GSSG/GSH) = -0.24 V, O₂ will be reduced (acting as oxidant) while GSH will be oxidized to GSSG (acting as reductant). This makes option A correct, identifying O₂ as the oxidant and GSH as the reductant. Option B incorrectly claims O₂ is oxidized despite having the higher E°. To identify oxidants and reductants: the species with higher E° is the oxidant (gets reduced), the species with lower E° is the reductant (gets oxidized).

Question 14

1. Ag(s) dissolves as Ag$^+$ is oxidized at the anode, and electrons flow from Ag to Zn
2. Zn(s) is oxidized at the anode, and electrons flow through the wire from Zn to Ag (correct answer)
3. Both electrodes undergo reduction because $E^\circ$ values are positive when added
4. Zn$^{2+}$ is reduced at the cathode, causing Zn(s) plating and a negative cell voltage

Explanation: This question tests the ability to identify spontaneous redox reactions in galvanic cells based on standard reduction potentials. In galvanic cells, the species with the more positive reduction potential undergoes reduction at the cathode, while the species with the more negative reduction potential undergoes oxidation at the anode. Since E°(Ag⁺/Ag) = +0.80 V > E°(Zn²⁺/Zn) = -0.76 V, Ag⁺ will be reduced to Ag(s) at the cathode while Zn(s) will be oxidized to Zn²⁺ at the anode. Electrons flow through the external wire from the anode (Zn) to the cathode (Ag), making option B correct. Option A incorrectly reverses the electron flow direction, while options C and D propose impossible scenarios for a spontaneous galvanic cell. To verify redox predictions, always compare reduction potentials: the higher E° species gets reduced, the lower E° species gets oxidized.

Question 15

1. Ag(s) is oxidized and Fe$^{2+}$ is reduced; electrons flow from Ag to Fe
2. Fe(s) is oxidized at the anode; Ag$^{+}$ is reduced at the cathode (correct answer)
3. Both half-reactions proceed as reductions because both $E^\circ$ values are given as reductions
4. The sign of $E^\circ$ indicates Fe$^{2+}$ must be reduced spontaneously in any cell

Explanation: The skill being tested is determining the spontaneous redox reaction and electrode roles in a galvanic cell using standard reduction potentials. In galvanic cells, the species with the more positive E° is reduced at the cathode, while the one with the less positive (or more negative) E° is oxidized at the anode. Here, Ag⁺ has E° = +0.80 V (higher than Fe's -0.44 V), so Ag⁺ is reduced at the cathode and Fe(s) is oxidized at the anode. This makes choice B correct, as it accurately identifies Fe(s) oxidation at the anode and Ag⁺ reduction at the cathode. Choice A fails because it incorrectly states Ag(s) is oxidized and electrons flow from Ag to Fe, whereas electrons flow from Fe (anode) to Ag (cathode). For similar redox scenarios, compare E° values to assign cathode (higher E°) and anode (lower E°). Verify spontaneity by ensuring E°_cell > 0, which drives electron flow from anode to cathode externally.

Question 16

1. The plating electrode is the anode because reduction occurs where metal deposits
2. The plating electrode is the cathode, and its mass increases as Cu(s) forms (correct answer)
3. Electrons are produced at the plating electrode and consumed at the opposite electrode
4. Cu$^{2+}$ is oxidized to Cu(s), so the solution becomes more concentrated in Cu$^{2+}$

Explanation: The skill being tested is distinguishing electrode roles and mass changes in an electrolytic cell for metal plating. In electrolytic cells, an external power source drives nonspontaneous reactions, with reduction (metal deposition) occurring at the cathode and oxidation at the anode. In this Cu plating setup, Cu²⁺ is reduced to Cu(s) at the plating electrode, which is the cathode, leading to mass increase as Cu deposits. This makes choice B correct, identifying the plating electrode as the cathode with increasing mass. Choice A fails because it misidentifies the plating electrode as the anode, but reduction (deposition) defines the cathode. For similar redox scenarios, confirm the cathode as the site of reduction in both galvanic and electrolytic cells. Verify direction by noting that in electrolysis, the power source pushes electrons to the cathode, forcing reduction.

Question 17

1. Fe$^{2+}$ is oxidized and Co$^{3+}$ is reduced; electrons flow from Fe$^{2+}$ to Co$^{3+}$ (correct answer)
2. Co$^{2+}$ is reduced to Co$^{3+}$ at the cathode because its potential is higher
3. Electrons flow from Co$^{3+}$ to Fe$^{3+}$ because both are oxidizing agents
4. No reaction occurs because both couples involve only ions, not metals

Explanation: 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.

Question 18

1. Ag(s) is oxidized and the Ag electrode loses mass
2. Cu(s) is oxidized and the Cu electrode loses mass (correct answer)
3. Electrons flow from Ag to Cu through the external circuit
4. The salt bridge must supply electrons to maintain charge neutrality

Explanation: The skill being tested is predicting mass changes and electron flow in a galvanic cell from standard reduction potentials. In galvanic cells, oxidation at the anode causes the anode electrode to lose mass if it's a metal, while reduction at the cathode may increase mass, with electrons flowing from anode to cathode. Here, Cu has E° = +0.34 V (lower than Ag's +0.80 V), so Cu(s) is oxidized at the anode (losing mass), and Ag⁺ is reduced at the cathode. This makes choice B correct, as Cu(s) oxidation leads to mass loss at the Cu electrode. Choice A fails because it incorrectly identifies Ag(s) as oxidized with mass loss, but Ag is the cathode. For similar redox scenarios, determine the anode by lower E° and note mass loss there if metal is oxidized. Verify electron flow from lower E° (anode) to higher E° (cathode).

Question 19

1. I$^-$ is oxidized and permanganate is reduced; electrons flow from the iodide half-cell to the permanganate half-cell (correct answer)
2. Mn$^{2+}$ is oxidized to MnO$_4^-$ spontaneously because its reduction potential is larger
3. Both species are reduced because both $E^\circ$ values are positive
4. Electrons flow from permanganate to iodide because permanganate has the higher $E^\circ$

Explanation: The skill being tested is identifying spontaneous redox direction and electron flow in a galvanic cell with non-metal half-cells using standard reduction potentials. In galvanic cells, the half-reaction with the higher E° occurs as reduction at the cathode, while the lower E° reverses to oxidation at the anode, with electrons flowing from anode to cathode. Here, MnO₄⁻ has E° = +1.51 V (higher than I₂'s +0.54 V), so I⁻ is oxidized to I₂ at the anode, and MnO₄⁻ is reduced at the cathode, with electrons from iodide to permanganate. This makes choice A correct, accurately stating I⁻ oxidation and permanganate reduction with correct flow. Choice D fails because it reverses electron flow; electrons flow to the higher E° (cathode). In similar redox scenarios, assign cathode to higher E° and anode to lower. Confirm direction by calculating E°_cell > 0 and noting anion oxidation if applicable.

Question 20

1. Cd$^{2+}$ is reduced at the cathode because it is a $2+$ ion
2. Cu(s) is oxidized at the anode because its reduction potential is more positive
3. Cd(s) is oxidized at the anode, and the Cd electrode mass decreases (correct answer)
4. The cell requires an external voltage because Cd has a negative $E^\circ$

Explanation: 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.