Electrochemical Cells and Redox Reactions (4C)
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MCAT Chemical and Physical Foundations of Biological Systems › Electrochemical Cells and Redox Reactions (4C)
A galvanic cell is prepared: Cd(s)|Cd$^{2+}$(1.0 M) || Ni$^{2+}$(1.0 M)|Ni(s). Standard reduction potentials: $E^\circ$(Cd$^{2+}$/Cd) = -0.40 V; $E^\circ$(Ni$^{2+}$/Ni) = -0.25 V. Which statement best reflects the redox principle illustrated?
Ni(s) is oxidized at the anode because its $E^\circ$ is higher
Cd$^{2+}$ is reduced at the anode because oxidation occurs at the cathode in galvanic cells
Cd(s) is oxidized at the anode; Ni$^{2+}$ is reduced at the cathode
Electrons flow from Ni to Cd because Cd has the more negative $E^\circ$
Explanation
This question tests the ability to predict spontaneous reactions in galvanic cells. The species with the more positive (less negative) reduction potential undergoes reduction at the cathode. Since E°(Ni²⁺/Ni) = -0.25 V > E°(Cd²⁺/Cd) = -0.40 V, Ni²⁺ will be reduced at the cathode while Cd(s) will be oxidized at the anode. This makes option A correct, with Cd(s) oxidation at the anode and Ni²⁺ reduction at the cathode, with electrons flowing from Cd to Ni. Option B incorrectly claims Ni is oxidized because of its "higher" E°, misunderstanding that less negative means more positive. For comparing negative E° values: -0.25 V > -0.40 V, so the -0.25 V species undergoes reduction.
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?
Mass increase indicates electrons are being produced there by reduction
The electrode gaining mass is the anode where metal atoms oxidize into solution
The electrode gaining mass must be connected to the negative terminal of a power supply
The electrode gaining mass is the cathode where metal ions are reduced and plate as solid
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).
A galvanic cell is built with Co(s)|Co$^{2+}$(1.0 M) and Br$_2$(l)|Br$^-$(1.0 M) on Pt. $E^\circ$(Co$^{2+}$/Co) = -0.28 V; $E^\circ$(Br$_2$/Br$^-$) = +1.07 V. Which outcome would be expected in this electrochemical cell under standard conditions?
Co(s) is oxidized at the anode and Br$_2$ is reduced to Br$^-$ at the cathode
Electrons flow from bromine to cobalt because halogens are strong reducing agents
Co$^{2+}$ is reduced at the cathode because it is the only cation present
Br$^-$ is reduced to Br$_2$ at the cathode and Co(s) is reduced at the anode
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.
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?
Cathodic processes cannot occur in electrolytic cells because they are nonspontaneous
OH$^-$ is oxidized at the cathode because it is negatively charged
Oxidation occurs at the cathode, producing electrons that reduce water
Reduction occurs at the cathode, consuming electrons supplied by the power source
Explanation
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 galvanic cell couples MnO$_4^-$/Mn$^{2+}$ and Fe$^{3+}$/Fe$^{2+}$ in acidic solution using Pt electrodes. Standard reduction potentials: $E^\circ$(MnO$_4^-$/Mn$^{2+}$) = +1.51 V; $E^\circ$(Fe$^{3+}$/Fe$^{2+}$) = +0.77 V. Which outcome would be expected under standard conditions?
Fe$^{3+}$ is reduced at the anode because anodes have higher potentials
Electrons flow from permanganate to iron because Mn contains more oxygen atoms
Mn$^{2+}$ is oxidized at the anode because it has the higher reduction potential
Fe$^{2+}$ is oxidized at the anode and MnO$_4^-$ is reduced at the cathode
Explanation
This question tests understanding of galvanic cells with polyatomic ions. The half-reaction with the more positive reduction potential proceeds as reduction at the cathode. Since E°(MnO₄⁻/Mn²⁺) = +1.51 V > E°(Fe³⁺/Fe²⁺) = +0.77 V, MnO₄⁻ will be reduced to Mn²⁺ at the cathode while Fe²⁺ will be oxidized to Fe³⁺ at the anode. This makes option A correct, with Fe²⁺ oxidation at the anode and MnO₄⁻ reduction at the cathode. Option B incorrectly suggests Mn²⁺ oxidation, which would require an even higher potential oxidant. For complex ions, apply the same principle: higher E° undergoes reduction, lower E° undergoes oxidation.
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?
The magnitude of the measured potential doubles because two leads are used
The sign of the measured cell potential reverses, but the direction of spontaneous electron flow in the cell does not
The anode and cathode swap physically because electron flow depends on lead placement
The cell becomes electrolytic and forces the nonspontaneous direction
Explanation
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.
A galvanic cell uses H$_2$(g, 1 atm)|H$^+$(1.0 M) as one half-cell and Ni$^{2+}$(1.0 M)|Ni(s) as the other. Given $E^\circ$(H$^+$/H$_2$) = 0.00 V and $E^\circ$(Ni$^{2+}$/Ni) = -0.25 V, which outcome would be expected under standard conditions?
H$_2$ is reduced at the cathode because its $E^\circ$ is higher than Ni$^{2+}$/Ni
Ni(s) is oxidized at the anode and H$^+$ is reduced to H$_2$ at the cathode
No reaction occurs because the standard hydrogen electrode cannot participate in galvanic cells
Ni(s) is reduced at the cathode and H$_2$ is oxidized at the anode
Explanation
This question tests understanding of the standard hydrogen electrode in galvanic cells. The standard hydrogen electrode (SHE) has E° = 0.00 V by definition, serving as the reference for all other potentials. Since E°(H⁺/H₂) = 0.00 V > E°(Ni²⁺/Ni) = -0.25 V, H⁺ will be reduced to H₂ at the cathode while Ni(s) will be oxidized to Ni²⁺ at the anode. This makes option C correct, with Ni(s) oxidation at the anode and H⁺ reduction at the cathode. Option A incorrectly suggests Ni reduction, while option B wrongly claims H₂ reduction. When SHE is involved, compare its E° = 0.00 V to the other half-cell: if the other E° is negative, that species gets oxidized.
In a heme-mimic experiment, Fe$^{3+}$ (aq) is reduced to Fe$^{2+}$ (aq) by ascorbate (AscH$^-$), which is oxidized to dehydroascorbate. Which statement best reflects the redox principle illustrated regarding oxidizing and reducing agents?
Ascorbate is the oxidizing agent because it loses electrons
Fe$^{2+}$ is the oxidizing agent because it has fewer positive charges
Both reactants are oxidizing agents because electron transfer is bidirectional
Fe$^{3+}$ is the oxidizing agent because it is reduced during the reaction
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).
A conceptual prediction is tested: In the galvanic cell Zn(s)|Zn$^{2+}$(1.0 M) || Cu$^{2+}$(1.0 M)|Cu(s), a student adds EDTA to the Cu half-cell, lowering free Cu$^{2+}$ by complexation. Which outcome would be expected for the cell voltage immediately after complexation (before significant mass change at electrodes)?
Electron flow reverses because EDTA is a strong oxidizing agent
Cell voltage increases because lowering free [Cu$^{2+}$] drives Cu$^{2+}$ reduction forward
Cell voltage decreases because the cathode reactant Cu$^{2+}$ becomes less available
Cell voltage is unchanged because complexation does not affect redox chemistry
Explanation
This question tests understanding of how complexation affects cell voltage through the Nernst equation. When EDTA complexes Cu²⁺, it dramatically reduces the free [Cu²⁺] available for the cathode reaction Cu²⁺ + 2e⁻ → Cu. According to the Nernst equation, decreasing the concentration of a reactant in a reduction reaction makes that reduction less favorable, lowering the cathode potential. Since cell voltage = E(cathode) - E(anode), and the anode potential remains constant, the overall cell voltage decreases. This makes option B correct, as the reduced availability of the cathode reactant Cu²⁺ decreases cell voltage. Option A incorrectly suggests voltage increase, while option C ignores the concentration effect. For complexation effects, remember: reducing free metal ion concentration at the cathode decreases cell voltage.
A galvanic cell is designed as a sensor with half-cells: Pb(s)|Pb$^{2+}$(1.0 M) and Ag$^+$(1.0 M)|Ag(s). $E^\circ$(Pb$^{2+}$/Pb) = -0.13 V; $E^\circ$(Ag$^+$/Ag) = +0.80 V. Based on the setup, which conclusion is most consistent with the electrochemical process?
Pb(s) is oxidized at the anode and Ag$^+$ is reduced at the cathode
Electrons flow from Ag to Pb because Ag has the higher reduction potential
Both Pb$^{2+}$ and Ag$^+$ are reduced because they are cations
Ag(s) is oxidized at the anode and Pb$^{2+}$ is reduced at the cathode
Explanation
This question tests the ability to predict spontaneous reactions in galvanic cells. The species with the more positive reduction potential undergoes reduction at the cathode, while the species with the more negative potential undergoes oxidation at the anode. Since E°(Ag⁺/Ag) = +0.80 V > E°(Pb²⁺/Pb) = -0.13 V, Ag⁺ will be reduced at the cathode while Pb(s) will be oxidized at the anode. This makes option B correct, with Pb(s) oxidation at the anode and Ag⁺ reduction at the cathode, and electrons flowing from Pb to Ag. Option A incorrectly identifies Ag as the anode, while option C reverses the electron flow direction. For galvanic cells, electrons always flow from the species with lower E° to the species with higher E°.