For a reaction to be spontaneous at low temperature, the signs of deltaHdegree a

Question

For a reaction to be spontaneous at low temperature, the signs of deltaHdegree and deltaSdegree must be and respectively. +,0 -, + +,- -,- +, + How many kilowatt-hours of electricity are used to produce 4.50 kg of magnesium in the electrolysis of molten MgCl_2 with an applied emf of 5.00 V? 0.0496 0.0201 24.8 12.4 A voltaic cell is constructed with two Zn^2+-Zn electrodes, where the half-reaction is The concentrations of zinc ion in the two compartments are 4.50 M and 1.11 times 10^-2 M, respectively. The cell emf is V.

Explanation / Answer

Dear Student,

1.Definition of entropy (S), the disorder of a system. It takes energy to force order on any system. If you have to pump energy in, then the reaction isn't very spontaneous, So, you want a positive entropy, which means that disorder is increasing.
Definition of enthalpy (heat.) If a reaction evolves heat, then it is a favorable reaction. However, if you have to put heat into it for the reaction to go, then it wasn't very spontaneous. Remember that if you have to add heat, the delta H is positive; if heat evolves on its own, the delta H is negative.
So, a reaction with positive S and a negative H is spontaneous.
This comes from Gibbs free energy relationship:
G = H -TS
If G is negative, then the reaction is spontaneous. To get a negative G, you would want the right side of the equation to be as negative as possible. H would ideally be negative, but S would be positive.

Therefore, option is B.

2. First let's see how many electrons we need. We produce 4500 grams of magnesium
with an average atomic weight of 24.3, so that is 4500/24.3, or 185.1 moles of Mg.
Each atom of Mg needs two electrons to reduce it to elementary magnesium so we
need 185.1 x 2, or 370.3 moles of electrons (also called faradays).
Each faraday contains Avogadro's number, or 6.0221 x 1023 electrons, so we need a
total of 370.3 x 6.0221 x 1023 or 2.229 x 1026 electrons.

Since one coulomb contains 6.28 x 1018 electrons, so we need (2.229 x 1026) / (6.28 x 1018) or 3.549 x 107 coulombs.

If we have a constant emf of 5.0 volts, that is a total of 17.74 x 107 volt-coulombs (3.549 x 107 x 5 ), or joules of energy. (177,400,000 joules).

Since 1 J equals one watt-second and there are 3600 seconds in an hour, one kilowatt hour
equals 3,600,000 joules.

In this case we use 177,400,000 joules so we use177,400,000 /3,600,000 or 49.27 kilowatt hours of energy.

Therefore, the option is D

Note: According to the formulas and the calculations the answer is 49.27 kilowatt hour. But in the options, option D i.e 49.6 is near to the answer. Hence consider this option as an answer.

3. Follow the nersnt equation

Ecell = Eocell - (0.0592/2)log( [Zn] / [Zn+2])

Ecell = (-0.763) - (0.0296) log (0.0111/4.5)

Ecell = -0.763 - (0.0296 x log (2.46 x 10-3))

Ecell = -0.763 - (0.0296 x (-2.6078))

Ecell = - 0.763 - (-0.0771)

Ecell = -0.763 + 0.0771

Ecell = - 0.6806 V

Note that according to the formula and calculations mention this is the correct answer which is not shown in the options.

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