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ella [17]
3 years ago
10

A student constructs a galvanic cell that has a strip of iron metal immersed in a solution of 0.1M Fe(NO3)2 as one half-cell and

a strip of aluminum metal immersed in a solution of 0.1M Al(NO3)3 as the other half-cell. The measured cell potential is less than zero when the positive terminal of the voltmeter is attached to the aluminum strip and the negative terminal is attached to the iron strip. Which half-cell is the anode (Fe or Al)?
Chemistry
1 answer:
MrMuchimi3 years ago
4 0

Answer:

Fe

Explanation:

The cell potential is:

ΔE°cell = E°red(red) - E°red(oxid)

Where, E°red(red) is the reduction potential of the substance that is reducing, and E°red(oxid) is the reduction potential of the substance that is oxidizing. For the reaction be spontaneous and happen, ΔE°cell > 0.

The reduction takes place in the cathode, which is the negative pole, and the oxidation in the anode, which is the positive pole. So, the electrons flow from the positive pole to the negative pole (anode to cathode).

Then, if the voltmeter measured a negative potential, it means that is was attached incorrectly. So, the anode is Fe.

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seropon [69]

Answer:

10 L

Explanation:

The only variables are pressure and volume, so we can use Boyle's Law:

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Data:

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Answer:

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4 years ago
NHCI(s) = NH3(g) + HCl(g) at 427°C, Kp = 4.8. Calculate Kc​
eduard

\tt Kp=Kc(RT)^{\Delta n}

NH₄Cl (s) ⇔ NH₃ (g) + HCl (g)

Δn = 2 (gas only)

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