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

Study 4c Electrochemical Cells Redox in MCAT Chemical and Physical Foundations of Biological Systems with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

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.

How to use these flashcards

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.

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

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QUESTION

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

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ANSWER

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

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Flashcard 1: 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 2: 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 3: 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 4: At 298 K298\,\text{K}298K, find log⁡K\log KlogK if n=1n=1n=1 and Ecell∘=0.0592 VE^\circ_{\text{cell}}=0.0592\,\text{V}Ecell∘​=0.0592V.

Answer: log⁡K=1\log K=1logK=1. From E∘=0.0592nlog⁡KE^\circ=\frac{0.0592}{n}\log KE∘=n0.0592​logK, rearrange to log⁡K=nE∘0.0592\log K=\frac{nE^\circ}{0.0592}logK=0.0592nE∘​, yielding 1 for given values.

Flashcard 5: What is the Faraday law relation between charge passed and moles of electrons transferred?

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

Flashcard 6: 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 7: 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 8: 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 9: 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 10: What is the standard cell potential formula in terms of cathode and anode reduction potentials?

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

Flashcard 11: If a half-reaction is reversed, how does its electrode potential change?

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

Flashcard 12: When balancing a redox reaction, how does multiplying a half-reaction by nnn affect E^ ?

Answer: E∘E^\circE∘ 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∘E^\circE∘.

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

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

Flashcard 14: What is the relationship between ΔG∘\Delta G^\circΔG∘ and Ecell∘E^\circ_{\text{cell}}Ecell∘​?

Answer: ΔG∘=−nFEcell∘\Delta G^\circ=-nFE^\circ_{\text{cell}}ΔG∘=−nFEcell∘​. This equation links free energy change to electrochemical work, where nnn is moles of electrons, FFF is Faraday's constant, and positive E∘E^\circE∘ yields negative ΔG∘\Delta G^\circΔG∘.

Flashcard 15: What equation relates Ecell∘E^\circ_{\text{cell}}Ecell∘​ to KKK at 298 K298\,\text{K}298K using base-10 logs?

Answer: Ecell∘=0.0592 Vnlog⁡KE^\circ_{\text{cell}}=\frac{0.0592\,\text{V}}{n}\log KEcell∘​=n0.0592V​logK. Derived from combining ΔG∘=−nFE∘\Delta G^\circ=-nFE^\circΔG∘=−nFE∘ and ΔG∘=−RTln⁡K\Delta G^\circ=-RT\ln KΔG∘=−RTlnK, simplified with base-10 log at 298 K.

Flashcard 16: What is the Nernst equation for a cell potential EEE in terms of E∘E^\circE∘, nnn, and QQQ at 298 K298\,\text{K}298K?

Answer: E=E∘−0.0592 Vnlog⁡QE=E^\circ-\frac{0.0592\,\text{V}}{n}\log QE=E∘−n0.0592V​logQ. The Nernst equation adjusts the standard potential for non-standard conditions using the reaction quotient QQQ at 298 K.

Flashcard 17: Identify the cathode half-reaction given E∘(A)=+0.20 VE^\circ(\text{A})=+0.20\,\text{V}E∘(A)=+0.20V and E∘(B)=−0.10 VE^\circ(\text{B})=-0.10\,\text{V}E∘(B)=−0.10V (both as reductions).

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

Flashcard 18: Find Ecell∘E^\circ_{\text{cell}}Ecell∘​ if Ecathode∘=+0.80 VE^\circ_{\text{cathode}}=+0.80\,\text{V}Ecathode∘​=+0.80V and Eanode∘=−0.20 VE^\circ_{\text{anode}}=-0.20\,\text{V}Eanode∘​=−0.20V (both reduction potentials).

Answer: Ecell∘=+1.00 VE^\circ_{\text{cell}}=+1.00\,\text{V}Ecell∘​=+1.00V. Using Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}Ecell∘​=Ecathode∘​−Eanode∘​, substitute values to get +0.80−(−0.20)=+1.00 V+0.80 - (-0.20)=+1.00\,\text{V}+0.80−(−0.20)=+1.00V.

Flashcard 19: At 298 K298\,\text{K}298K, find EEE if E∘=0.30 VE^\circ=0.30\,\text{V}E∘=0.30V, n=2n=2n=2, and Q=10Q=10Q=10.

Answer: E≈0.27 VE\approx^0.27\,\text{V}E≈0.27V. Using the Nernst equation E=E∘−0.0592nlog⁡QE=E^\circ-\frac{0.0592}{n}\log QE=E∘−n0.0592​logQ, substitute values to yield 0.30−0.0296×1≈0.27 V0.30 - 0.0296 \times 1 \approx 0.27\,\text{V}0.30−0.0296×1≈0.27V.

Flashcard 20: 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 21: 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.