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

  1. In a galvanic cell, which way do cations and anions migrate through the salt bridge?
    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.
  2. In an electrolytic cell, what is the sign of the anode and the cathode?
    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.
  3. What is the reducing agent in a redox reaction, in terms of what happens to it?
    The reducing agent is oxidized (loses electrons). The reducing agent donates electrons to the oxidizing agent, thereby undergoing oxidation itself in the reaction.
  4. What is the oxidizing agent in a redox reaction, in terms of what happens to it?
    The oxidizing agent is reduced (gains electrons). The oxidizing agent accepts electrons from the reducing agent, thereby undergoing reduction itself in the reaction.
  5. Which electrode is the anode in any electrochemical cell, defined by the process occurring there?
    Anode = site of oxidation. By definition, the anode is where oxidation occurs, as electrons are released during the loss of electrons.
  6. What is the standard cell potential formula in terms of cathode and anode reduction potentials?
    Ecell∘=Ecathode∘−Eanode∘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.
  7. What equation relates Ecell∘E^\circ_{\text{cell}} to KK at 298 K298\,\text{K} using base-10 logs?
    Ecell∘=0.0592 Vnlog⁡KE^\circ_{\text{cell}}=\frac{0.0592\,\text{V}}{n}\log K. Derived from combining ΔG∘=−nFE∘\Delta G^\circ=-nFE^\circ and ΔG∘=−RTln⁡K\Delta G^\circ=-RT\ln K, simplified with base-10 log at 298 K.
  8. When balancing a redox reaction, how does multiplying a half-reaction by nn affect $E^ $?
    E∘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∘E^\circ.
  9. If a half-reaction is reversed, how does its electrode potential change?
    The sign of E∘E^\circ reverses. Reversing a half-reaction changes it from reduction to oxidation, which negates the potential value.
  10. Find Ecell∘E^\circ_{\text{cell}} if Ecathode∘=+0.80 VE^\circ_{\text{cathode}}=+0.80\,\text{V} and Eanode∘=−0.20 VE^\circ_{\text{anode}}=-0.20\,\text{V} (both reduction potentials).
    Ecell∘=+1.00 VE^\circ_{\text{cell}}=+1.00\,\text{V}. Using Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}, substitute values to get +0.80−(−0.20)=+1.00 V+0.80 - (-0.20)=+1.00\,\text{V}.
  11. At 298 K298\,\text{K}, find log⁡K\log K if n=1n=1 and Ecell∘=0.0592 VE^\circ_{\text{cell}}=0.0592\,\text{V}.
    log⁡K=1\log K=1. From E∘=0.0592nlog⁡KE^\circ=\frac{0.0592}{n}\log K, rearrange to log⁡K=nE∘0.0592\log K=\frac{nE^\circ}{0.0592}, yielding 1 for given values.
  12. What is the Faraday law relation between charge passed and moles of electrons transferred?
    ne−=QFn_{e^-}=\frac{Q}{F}. Faraday's first law states that the moles of electrons transferred equal the total charge divided by Faraday's constant.
  13. What is the spontaneity criterion relating Ecell∘E^\circ_{\text{cell}} to a galvanic reaction?
    Spontaneous if Ecell∘>0E^\circ_{\text{cell}}>0. A positive cell potential indicates a favorable driving force for the reaction, making it spontaneous under standard conditions.
  14. In a galvanic (voltaic) cell, what is the sign of the anode and the cathode?
    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.
  15. What is the direction of electron flow in the external circuit of any electrochemical cell?
    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.
  16. Which electrode is the cathode in any electrochemical cell, defined by the process occurring there?
    Cathode = site of reduction. By definition, the cathode is where reduction occurs, as electrons are gained during the process.
  17. What is the definition of standard reduction potential E^ for a half-reaction?
    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.
  18. At 298 K298\,\text{K}, find EE if E∘=0.30 VE^\circ=0.30\,\text{V}, n=2n=2, and Q=10Q=10.
    E≈0.27 VE\approx^0.27\,\text{V}. Using the Nernst equation E=E∘−0.0592nlog⁡QE=E^\circ-\frac{0.0592}{n}\log Q, substitute values to yield 0.30−0.0296×1≈0.27 V0.30 - 0.0296 \times 1 \approx 0.27\,\text{V}.
  19. Identify the cathode half-reaction given E∘(A)=+0.20 VE^\circ(\text{A})=+0.20\,\text{V} and E∘(B)=−0.10 VE^\circ(\text{B})=-0.10\,\text{V} (both as reductions).
    Cathode is A (more positive E∘E^\circ). The half-reaction with the higher (more positive) reduction potential has greater tendency to be reduced, thus serving as the cathode.
  20. What is the Nernst equation for a cell potential EE in terms of E∘E^\circ, nn, and QQ at 298 K298\,\text{K}?
    E=E∘−0.0592 Vnlog⁡QE=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 QQ at 298 K.
  21. What is the relationship between ΔG∘\Delta G^\circ and Ecell∘E^\circ_{\text{cell}}?
    ΔG∘=−nFEcell∘\Delta G^\circ=-nFE^\circ_{\text{cell}}. This equation links free energy change to electrochemical work, where nn is moles of electrons, FF is Faraday's constant, and positive E∘E^\circ yields negative ΔG∘\Delta G^\circ.

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