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

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This deck focuses on 4c Electrochemical Cells Redox, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: In a galvanic cell, which way do cations and anions migrate through the salt bridge?

Answer: Cations to cathode; anions to anode. Cations move to the cathode to balance negative charge buildup from reduction, while anions move to the anode to balance positive charge from oxidation.

Flashcard 2: In an electrolytic cell, what is the sign of the anode and the cathode?

Answer: Anode is positive; cathode is negative. In electrolytic cells, the external power source makes the anode positive to attract anions for oxidation and the cathode negative for reduction.

Flashcard 3: What is the reducing agent in a redox reaction, in terms of what happens to it?

Answer: The reducing agent is oxidized (loses electrons). The reducing agent donates electrons to the oxidizing agent, thereby undergoing oxidation itself in the reaction.

Flashcard 4: What is the oxidizing agent in a redox reaction, in terms of what happens to it?

Answer: The oxidizing agent is reduced (gains electrons). The oxidizing agent accepts electrons from the reducing agent, thereby undergoing reduction itself in the reaction.

Flashcard 5: Which electrode is the anode in any electrochemical cell, defined by the process occurring there?

Answer: Anode = site of oxidation. By definition, the anode is where oxidation occurs, as electrons are released during the loss of electrons.

Flashcard 6: What is the standard cell potential formula in terms of cathode and anode reduction potentials?

Answer: $E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$. The cell potential is calculated by subtracting the anode's reduction potential from the cathode's, accounting for the oxidation at the anode.

Flashcard 7: What equation relates $E^\circ_{\text{cell}}$ to $K$ at $298\,\text{K}$ using base-10 logs?

Answer: $E^\circ_{\text{cell}}=\frac{0.0592\,\text{V}}{n}\log K$. Derived from combining $\Delta G^\circ=-nFE^\circ$ and $\Delta G^\circ=-RT\ln K$, simplified with base-10 log at 298 K.

Flashcard 8: When balancing a redox reaction, how does multiplying a half-reaction by $n$ affect E^?

Answer: $E^\circ$ does not change when coefficients are scaled. Electrode potentials are intensive properties, independent of the amount of substance, so scaling coefficients does not alter $E^\circ$.

Flashcard 9: If a half-reaction is reversed, how does its electrode potential change?

Answer: The sign of $E^\circ$ reverses. Reversing a half-reaction changes it from reduction to oxidation, which negates the potential value.

Flashcard 10: Find $E^\circ_{\text{cell}}$ if $E^\circ_{\text{cathode}}=+0.80\,\text{V}$ and $E^\circ_{\text{anode}}=-0.20\,\text{V}$ (both reduction potentials).

Answer: $E^\circ_{\text{cell}}=+1.00\,\text{V}$. Using $E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$, substitute values to get $+0.80 - (-0.20)=+1.00\,\text{V}$.

Flashcard 11: At $298\,\text{K}$, find $\log K$ if $n=1$ and $E^\circ_{\text{cell}}=0.0592\,\text{V}$.

Answer: $\log K=1$. From $E^\circ=\frac{0.0592}{n}\log K$, rearrange to $\log K=\frac{nE^\circ}{0.0592}$, yielding 1 for given values.

Flashcard 12: What is the Faraday law relation between charge passed and moles of electrons transferred?

Answer: $n_{e^-}=\frac{Q}{F}$. Faraday's first law states that the moles of electrons transferred equal the total charge divided by Faraday's constant.

Flashcard 13: What is the spontaneity criterion relating $E^\circ_{\text{cell}}$ to a galvanic reaction?

Answer: Spontaneous if $E^\circ_{\text{cell}}>0$. A positive cell potential indicates a favorable driving force for the reaction, making it spontaneous under standard conditions.

Flashcard 14: In a galvanic (voltaic) cell, what is the sign of the anode and the cathode?

Answer: Anode is negative; cathode is positive. In galvanic cells, the anode accumulates negative charge from electron release, while the cathode is positive from electron consumption.

Flashcard 15: What is the direction of electron flow in the external circuit of any electrochemical cell?

Answer: Electrons flow from anode to cathode. Electrons are produced at the anode via oxidation and consumed at the cathode via reduction, driving flow through the external circuit.

Flashcard 16: Which electrode is the cathode in any electrochemical cell, defined by the process occurring there?

Answer: Cathode = site of reduction. By definition, the cathode is where reduction occurs, as electrons are gained during the process.

Flashcard 17: What is the definition of standard reduction potential E^ for a half-reaction?

Answer: Potential for reduction under standard conditions vs SHE. Standard reduction potential measures the tendency of a species to gain electrons relative to the standard hydrogen electrode under standard conditions.

Flashcard 18: At $298\,\text{K}$, find $E$ if $E^\circ=0.30\,\text{V}$, $n=2$, and $Q=10$.

Answer: $E\approx^0.27\,\text{V}$. Using the Nernst equation $E=E^\circ-\frac{0.0592}{n}\log Q$, substitute values to yield $0.30 - 0.0296 \times 1 \approx 0.27\,\text{V}$.

Flashcard 19: Identify the cathode half-reaction given $E^\circ(\text{A})=+0.20\,\text{V}$ and $E^\circ(\text{B})=-0.10\,\text{V}$ (both as reductions).

Answer: Cathode is A (more positive $E^\circ$). The half-reaction with the higher (more positive) reduction potential has greater tendency to be reduced, thus serving as the cathode.

Flashcard 20: What is the Nernst equation for a cell potential $E$ in terms of $E^\circ$, $n$, and $Q$ at $298\,\text{K}$?

Answer: $E=E^\circ-\frac{0.0592\,\text{V}}{n}\log Q$. The Nernst equation adjusts the standard potential for non-standard conditions using the reaction quotient $Q$ at 298 K.

Flashcard 21: What is the relationship between $\Delta G^\circ$ and $E^\circ_{\text{cell}}$?

Answer: $\Delta G^\circ=-nFE^\circ_{\text{cell}}$. This equation links free energy change to electrochemical work, where $n$ is moles of electrons, $F$ is Faraday's constant, and positive $E^\circ$ yields negative $\Delta G^\circ$.