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2 years ago

Complete the following sentence: The cathode in a voltaic cell is the electrode _____________.

Chemistry

1 answer:

Sergio [31]2 years ago

7 0

Answer:

to which cations from the salt bridge migrate

Explanation:

A voltaic cell is an electrochemical cell that uses spontaneous redox reactions to generate electricity. It's composed of a cathode, an anode, and a salt bridge.

In cathode, the substance is gaining electrons, so it's reducing, in the anode, the substance is losing electrons, so it's oxidating. The flow of electrons is from the anode to the cathode.

The salt bridge is a bond between the cathode and the anode. When the redox reaction takes place, the substances produce its ions, so the solution is no more neutral. The salt bridge allows the solutions to become neutral and the redox reaction continues.

So, the cathode produces anions, which goes to the anode, and the anode produces cations, which goes to the cathode. Then, the cathode n a voltaic cell is the electrode to which cations from salt bridge migrate and where the reduction takes place.

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Enter the chemical equation 2H+(aq)+S2−(aq)→H2S(g). Express your answer as a chemical equation.

amm1812

Answer : The complete chemical equation is,

$2H^+(aq)+S^{2-}(aq)\rightarrow H_2S(aq)$

Explanation :

As we know that, in a chemical equation the reacting species present on left side and the product formed present on right side and a right arrow inserted between the reactants and product that show a chemical reaction taking place.

In the chemical reaction, the phases of the substances are also included and subscripts and superscripts are also used for the numbers.

For the given chemical reaction, the balanced chemical equation including the phases, is given by:

$2H^+(aq)+S^{2-}(aq)\rightarrow H_2S(aq)$

4 0

1 year ago

Read 2 more answers

3. A 0.500 g sample of nitrogen gas combines with 1.140 g of oxygen gas to form NO2. If the atomic mass of oxygen is 16.000, cal

DedPeter [7]

Answer;

= 18.24

Explanation;

The ratio of N and O in the formula NO2 IS 1:2

Mass of nitrogen gas is 0.500 g

Moles of nitrogen will be;

= 0.500/16 = 0.03125 moles

Therefore;

The moles of Oxygen from the ratio will be;

= 0.03125 × 2 = 0.0625 moles

But; 0.0625 moles is equal to 1.140 g of Oxygen

The atomic number (mass in 1 mole) will be;

= 1.140 /0.0625

= 18.24

Thus the atomic number of Oxygen from the data is 18.24

6 0

2 years ago

Read 2 more answers

How many milliliters of 0.02 M HCl are needed to react completely with 100 mL of 0.01 M NaOH?

serious [3.7K]

The reaction of HCl and NaOH is HCl + NaOH = NaCl + H2O. So the mole number of HCl and NaOH is equal. So the volume of HCl =0.01*0.1/0.02=0.05 L =50 ml. So the answer is D).

3 0

2 years ago

Read 2 more answers

Which of the following species contains manganese with the highest oxidation number?

ioda

In NaMnO₄, Mn has the highest oxidation number.

The question is incomplete, the complete question is;

Which of the following species contains manganese with the highest oxidation number?

A) Mn

B) MnF₂

C) Mn₃(PO₄)₂

D) MnCl₄

E) NaMnO₄

In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.

1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.

2) For MnF₂;

Mn has an oxidation number of +2

3) For Mn₃(PO₄)₂

Mn has an oxidation number of +2

4) For MnCl₄

Mn has an oxidation number of +4

5) For NaMnO₄

Mn has an oxidation number of +7

Hence in NaMnO₄, Mn has the highest oxidation number.

Learn more: brainly.com/question/10079361

7 0

1 year ago

I NEED HELP ASAP, WILL MARK BRAINLEST!

Andre45 [30]

Answer:

1. 90%

2. 217.4 g O₂

3. 95.0%

4. Trial 2 ratios

Explanation:

Original: SiCl₄ + O₂ → SiO₂ + Cl₂

Balanced: SiCl₄ + O₂ → SiO₂ + 2Cl₂

Trial SiCl₄ O₂ SiO₂

1 120 g 240 g 38.2 g

2 75 g 50 g 25.2 g

Percentage yield for trial 1

We need to get actual yield (38.2 g) and theoretical yield, in grams.

Mass to moles:

molar mass SiCl₄: 28.09 + 4(35.45) = 169.9 g/mol

120 g SiCl₄ x 1 mol/169.9 g = .706 mol SiCl₄

Moles to moles:

For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

.706 mol SiCl₄ = .706 mol SiO₂

Moles to mass:

molar mass SiO₂: 28.09 + 2(16.00) = 60.09 g/mol

.706 mol SiO₂ x 60.09g/mol = 42.44 g SiO₂

Theoretical yield:

actual/theoretical x 100

38.2 / 42.44 = .900 = 90.0% yield

Leftover reactant for trial 1

We know oxygen is the excess reactant.

Mass to moles:

molar mass O₂ = 32.00 g/mol

240 g O₂ x 1 mol/32.00 g = 7.5 mol O₂

We used .706 mol SiO₂, so we also used .706 mol O₂.

7.5 - .706 = 6.8 moles left over

Moles to mass:

6.8 mol O₂ x 32.00g/mol = 217.4 g O₂

Percentage yield for trial 2

Mass to moles:

molar mass SiCl₄: 169.9 g/mol

75 g SiCl₄ x 1 mol/169.9 g = .441 mol SiCl₄

Moles to moles:

For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

.441 mol SiCl₄ = .441 mol SiO₂

Moles to mass:

molar mass SiO₂: 60.09 g/mol

.441 mol SiO₂ x 60.09g/mol = 26.5 g SiO₂

Theoretical yield:

actual/theoretical x 100

25.2 / 26.5 = .950 = 95.0% yield

Because the percentage yield of trial 2 is higher than that of trial 1, we know that the ratio of reactants in trial 2 is more efficient! We got a result closer to our theoretical yield.

6 0

2 years ago