Galvanic Cells and Electrode Potentials: Quiz

Galvanic Cells and Electrode Potentials: Quiz

7 questions

1. Question 1EasyIn a galvanic cell, where does oxidation take place?

   AAt the cathodeBAt the anodeCIn the salt bridgeDIn the voltmeter
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   Answer: At the anode

   Oxidation always occurs at the anode and reduction at the cathode.

2. Question 2MediumUsing E°(Ag⁺/Ag) = +0.80 V and E°(Zn²⁺/Zn) = −0.76 V, what is E°cell for a zinc–silver cell?

   A+0.04 VB+1.56 VC+2.36 VD−1.56 V
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   Answer: +1.56 V

   Silver is reduced (cathode): E°cell = +0.80 V − (−0.76 V) = +1.56 V. Doubling the silver value to 1.60 V would be wrong; the answer 2.36 V does that.

3. Question 3MediumWhich species is the strongest oxidizing agent: Ag⁺ (+0.80 V), Cu²⁺ (+0.34 V), Zn²⁺ (−0.76 V) or Mg²⁺ (−2.37 V)?

   AAg⁺BCu²⁺CZn²⁺DMg²⁺
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   Answer: Ag⁺

   The most positive reduction potential means the species is most easily reduced, so it is the strongest oxidizing agent.

4. Question 4HardWill copper metal react with 1 M Zn²⁺ solution? (Cu²⁺/Cu +0.34 V, Zn²⁺/Zn −0.76 V)

   AYes, E°cell = +1.10 VBNo, E°cell = −1.10 VCYes, E°cell = −0.42 VDNo, E°cell = 0 V
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   Answer: No, E°cell = −1.10 V

   For Cu + Zn²⁺ → Cu²⁺ + Zn, Zn²⁺ would be the cathode: E°cell = −0.76 V − (+0.34 V) = −1.10 V. Negative means not spontaneous.

5. Question 5MediumWhat is ΔG° for the Daniell cell (E°cell = +1.10 V, n = 2)?

   A−106 kJB−212 kJC+212 kJD−0.212 kJ
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   Answer: −212 kJ

   ΔG° = −(2 mol)(96 485 C/mol)(1.10 V) = −2.12 × 10⁵ J = −212 kJ (C × V = J). The answer −106 kJ uses n = 1.

6. Question 6EasyWhat is the purpose of the salt bridge?

   ATo carry electrons between the half-cellsBTo let ions flow and keep the solutions neutralCTo increase the cell voltageDTo act as a catalyst
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   Answer: To let ions flow and keep the solutions neutral

   Electrons travel through the wire; ions move through the salt bridge. Without it, charge would build up and the current would stop.

7. Question 7MediumA current of 2.00 A flows for 30.0 min. How much charge passes?

   A60.0 CB3600 CC15.0 CD1.80 × 10⁵ C
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   Answer: 3600 C

   Q = It = 2.00 C/s × (30.0 min × 60 s/min) = 2.00 C/s × 1800 s = 3600 C. The answer 60.0 C forgets to convert minutes to seconds.