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3 years ago

When an aqueous solution of sodium sulfate is electrolyzed, a gas is observed to form at the anode. the gas is?

1 answer:

8 0

Electrolysis of Aqueous SolutionsIn the electrolysis of aqueous solutions only one ion is involved in the selective discharge of ions at each electrode during the electrolysis. The ion which is selected for discharge at an electrode depends on a number of factors, includingPosition of the ions in the electrochemical series,For positive ions, the facility of discharge decrease in going from those least electropositive to those most electropositive. For example, if both copper and hydrogen ions are present in solution, it will be the copper ions which take electrons from the cathode to become copper atoms.For negative ions, the ease of discharge decrease in going from those least electronegative to those most electronegative.Concentration of the Ions in SolutionIrrespective of the position of the ions in the electrochemical series, there is a tendency to promote the discharge of the most concentrated ion present. For example, in concentrated sodium chloride solution (i.e. brine) , the two cations present are the chlorine ion and the hydroxyl ion. Although the hydroxyl ion is more easily oxidised than the chlorine ion, it is the chlorine ion which will be discharged because its concentration is much greater than that of the hydroxyl ion.Nature of the ElectrodeThis is not as important as either of the other two factors, except in certain cases. For example in the electrolysis of molten sodium chloride using a mercury cathode, sodium ions are discharged in preference to hydrogen ions which are lower in the series.

Electrolysis of an Aqueous Copper Sulphate Solution using Copper Electrodes

The electrolysis of an aqueous solution of copper sulphate using copper electrodes (i.e. using active electrodes) results in transfer of copper metal from the anode to the cathode during electrolysis. The copper sulphate is ionised in aqueous solution.

CuSO4 ==> Cu(++) + SO4(-.-)

The positively charged copper ions migrate to the cathode, where each gains two electrons to become copper atoms that are deposited on the cathode.

Cu(++) + 2e(-) ==> Cu

At the anode, each copper atom loses two electrons to become copper ions, which go into solution.

Cu ==> Cu(++) + 2e(-)

The sulphate ion does not take part in the reaction and the concentration of the copper sulphate in solution does not change. The reaction is completed when the anode is completely eaten away. This process is used in electroplating.

Electrolysis of an Aqueous Solution of Sodium SulphateThe electrolysis of an aqueous solution of sodium sulphate using inert electrodes produces hydrogen at the cathode and oxygen at the anode and a neutral solution of sodium sulphate remains unaltered by the electrolysis.

Cathode Reaction : 4 H2O + 4 e(-) ==> 2 H2 + 4 OH(-)

Anode Reaction : 2 H2O ==> O2 + 4 H(+) + 4 e(-)

The overall cell reaction is : 6 H2O ==> 2 H2 + O2 +4 H(+) +4 OH(-)

If the reaction is carried out in a Hofmann Voltammeter, with some universal indicator in the solution, it will be noticed that around the cathode the solution becomes alkaline and around that anode the solution becomes acidic. This is explained as follows :

At the cathode :Hydrogen ions are being removed from solution, thereby leaving an excess of hydroxyl ions which makes the solution alkaline, andAt the anode :Hydroxyl ions are being removed, so leaving an excess of hydrogen ions which makes the solution acidic.

Electrolysis of a solution of dilute Sulphuric AcidThe electrolysis of an aqueous solution of dilute sulphuric acid is often carried out in a Hofmann Voltammeter, an apparatus in which the gases evolved at the anode and cathode can be collected in separate graduated tubes. When the solution is electrolyzed hydrogen is produced at the cathode and oxygen at the anode. These gases can be shown to be present in a 2 to 1 ratio and result from the electrolysis of water under acidic conditions.

Sulphuric acid is a strong electrolyte is fully dissociated in aqueous solution.

H2SO4 ==> 2 H(+) + SO4(2 -)

Water is a weak electrolyte and is only slightly dissociated

H2O ==> H(+) + OH(-)

During electrolysis, the hydrogen ions migrates towards the cathode, and are discharged there (i.e. they gain an electron and are converted to hydrogen gas).

2 H(+) + 2 e(-) ==> H2

At the anode the concentration of hydroxyl ions is too low to maintain a reaction and the sulphate ions are not oxidized but remain on in solution at the end. Water molecules must be the species reacting at the anode.

2 H2O ==> O2 + 4 H(+) + 4 e(-)

The overall reaction is

Cathode Reaction :2 H(+) + 2e(-) ==> H2 4 H(+) + 4e(-) ==> 2H2 Anode Reaction :2 H2O ==> O2 + 4 H(+) + 4 e(-) Overall Cell Reaction:4 H(+) + 2 H2O ==> 2 H2 + O2 + 4 H(+)

For every hydrogen ions discharged at the anode, another hydrogen ion is formed at the cathode. The net result is that the concentration of the sulphuric acid remains constant and this electrolysis consists of the decomposition of water with the overall reaction

2H2O ==> 2H2 + O2

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Consider the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g). If 0.02 moles of PCl5, 0.04 moles of PCl3, and 0.08 moles of Cl2 are combined

Furkat [3]

Answer:

The reaction quotient (Q) before the reaction is 0.32

Explanation:

Being the reaction:

aA + bB ⇔ cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.

The concentration will be calculated by:

$Concentration=\frac{number of moles of solute}{Volume}$

You know the reaction:

PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).

So:

$Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}$

The concentrations are:

[PCl₃]= $\frac{0.04 moles}{0.5 L} =0.08 \frac{moles}{L}$

[Cl₂]= $\frac{0.08 moles}{0.5 L} =0.16 \frac{moles}{L}$

[PCl₅]= $\frac{0.02 moles}{0.5 L} =0.04 \frac{moles}{L}$

Replacing:

$Q=\frac{0.08*0.16}{0.04}$

Solving:

Q= 0.32

The reaction quotient (Q) before the reaction is 0.32

4 0

3 years ago

In contracting skeletal muscle, a sudden elevation of cytosolic Ca2 concentration will result in: Inactivation of phosphorylase

skad [1K]

Answer:

ACTIVATION OF PHOSPHORYLASE KINASE AND ALLOSTERICALLY ACTIVATION OF PHOSPHORYLASE KINASE B

Explanation:

In a contracting skeletal muscle, there is a rapid need of ATP by the muscle cell. The energy need is obtained by the degradation of glycogen into glucose which then enters glycolysis. Muscle contraction causes muscle depolarization in which there is the rapid influx of calcium ions from the sarcoplasmic reticulum into the sacroplasm of the myocytes. This in turn causes the binding of calcium ion with calmodulin which thenb activates phosphorylase kinase from which it allosterically activate the b form of the enzyme needed for the conversion of glycogen to glucose. The other options do not occur as a result of increase in cytolic calcium concentration.

4 0

3 years ago

Classify each process by its individual effect on the entropy of the universe, s. hydrogen peroxide slowly decomposing into wate

Jobisdone [24]

Hydrogen Peroxide slowly decomposing into water and oxygen increases the entropy of the universe. Any reaction of a system always increases the degree of the orderliness of the universe. Decomposition is one of the best examples of increasing the entropy. This is when substance is broken down further to elements or other compounds.

4 0

3 years ago

A chemist prepares a solution of silver(I) nitrate by measuring out of silver(I) nitrate into a volumetric flask and filling the

leva [86]

Answer:

3.33 M

Explanation:

It seems your question is incomplete, however, that same fragment has been found somewhere else in the web:

" A chemist prepares a solution of silver nitrate (AgNO3) by measuring out 85.g of silver nitrate into a 150.mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's silver nitrate solution. Be sure your answer has the correct number of significant digits. "

In this case, first we calculate the moles of AgNO₃, using its molecular weight:

85.0 g AgNO₃ ÷ 169.87 g/mol = 0.500 mol AgNO₃

Then we convert the 150 mL of the volumetric flask into L:

150 / 1000 = 0.150 L

Finally we divide the moles by the volume:

0.500 mol AgNO₃ / 0.150 L = 3.33 M

4 0

3 years ago

Why bond angle of H2O is maximum then OF2??

Nataly_w [17]

It's lone a little distinction (103 degrees versus 104 degrees in water), and I trust the standard rationalization is that since F is more electronegative than H, the electrons in the O-F bond invest more energy far from the O (and near the F) than the electrons in the O-H bond. That moves the powerful focal point of the unpleasant constrain between the bonding sets far from the O, and thus far from each other. So the shock between the bonding sets is marginally less, while the repugnance between the solitary matches on the O is the same - the outcome is the edge between the bonds is somewhat less.

8 0

4 years ago