Questions Galvanic Cell Metal and Solution in Cathode Half-cell Cell Potential(v) Metal and Solution in Anode...

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Galvanic Cell Metal and Solution in Cathode Half-cell Cell Potential(v) Metal and Solution in Anode Half-cell Black Wire (-)

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5. Based on your information above write the half reactions occurring in each half-cell. Ensure you are writing the correct o

15. Using the standard reduction potential that you calculated in question #7 for the known metal you used, calculate the sta

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Answer 1

Answer #1

Q-5

Oxidation takes place at anode and reduction takes place at cathode.

Galvanic cell	Reaction occuring at anode	Reaction occuring at cathode
#1	H₂ (g) -------> 2H⁺ (aq) + 2e^-	Mg²⁺ (aq) + 2e^- ---------> Mg(s)
#2	H₂ (g) -------> 2H⁺ (aq) + 2e^-	Pb²⁺ (aq) + 2e^- ---------> Pb(s)
#3	H₂ (g) -------> 2H⁺ (aq) + 2e^-	Ni²⁺ (aq) + 2e^- ---------> Ni(s)

Q-6

Adding the two half cells reaction, we get the overall balanced reaction.

Galvanic cell	Overall balanced reaction
#1	H₂ (g) + Mg²⁺ (aq) <-------> 2H⁺ (aq) + Mg(s)
#2	H₂ (g) + Pb²⁺ (aq) <-------> 2H⁺ (aq) + Pb(s)
#3	H₂ (g) + Ni²⁺ (aq) <-------> 2H⁺ (aq) + Ni(s)

Q-7

Given, E⁰_red = 0 V for hydrogen

Assuming that the cell potential is measured under standard conditions

E⁰cell = E⁰_red(cathode) - E⁰_red(anode) => E⁰cell = E⁰_red(cathode)

Metal #1

E⁰cell = E⁰_red(cathode) = -2.37 V

Metal #2

E⁰cell = E⁰_red(cathode) = -0.143 V

Metal #3

E⁰cell = E⁰_red(cathode) = -0.249 V

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Answer 2

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