1)Determine the number of electrons formally transferred in the balanced equatio

Question

1)Determine the number of electrons formally transferred in the balanced equation . 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

2)Given the balanced equation, for the species that forms as a result of accepting electrons, enter the symbol and the final oxidation state of the specific element that was reduced. Use arabic numerals for oxidation state magnitude. Example: Element: H Oxidation state: +1

Sn(s) + 2Ag+(aq) + 2ClO4-(aq) 2Ag(s) + Sn2+(aq) + 2ClO4-(aq)

Element:

Oxidation state:

3)Given the unbalanced equation, enter the formula and state of the species that is oxidized. If more than one answer exists, separate by a comma. Cd(s) + Ag+(aq) Ag(s) + Cd2+(aq)

4)Given the balanced equation, enter the formula and state of the product that formed as a result of its precursor donating electrons. If more than one answer exists, separate by a comma. 3Zn(s) + 2Au3+(aq) + 6Cl-(aq) 2Au(s) + 3Zn2+(aq) + 6Cl-(aq)

5)Given the unbalanced equation, for the species that is the reduction product, enter the symbol and the final oxidation state of the specific element that has a lower oxidation state. Use arabic numerals for oxidation state magnitude. Example: Element: Sb Oxidation state: 0

Ba(s) + H2O(l) Ba2+(aq) + OH-(aq) + H2(g)

Element:

Oxidation state:

6)Given the balanced equation, for the species that undergoes reduction, enter the symbol and the initial oxidation state of the specific element that will go to a lower oxidation state. Enter sign followed by Arabic numeral for oxidation state magnitude. Use 0 if the oxidation state is zero. Example: Element: S Oxidation state: -2

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

Element:

Oxidation state:

Explanation / Answer

1. In the equation:

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

Na is oxidised and H is reduced.

the reduction half reaction is:

2H2O + 2e- -------------> H2 + 2OH-

The oxidation half reaction is:

2Na ----------> 2Na+ + 2e-

2Na+ + 2OH- ----------> 2NaOH

Combining these equations, we get :

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

So, number of electrons formally transferred = 2

2. In the equation:

Sn(s) + 2Ag+(aq) + 2ClO4-(aq) 2Ag(s) + Sn2+(aq) + 2ClO4-(aq)

In the reaction, Sn (0) is oxidised to Sn(II)

Ag(I) is reduced to Ag (0)

So, Element: Ag

Oxidation state: 0

3. In the equation:

Cd(s) + Ag+(aq) Ag(s) + Cd2+(aq)

Reduction reaction:

Ag+ + e- ----------> Ag

Oxidation reaction:

Cd -----------> Cd2+ + 2e-

In reduction reaction, one electron is added and in oxidation reaction, 2 electrons are removed. To balance, we need to multiply the reduction reaction with 2.

So, Final balanced equation is:

Cd(s) + 2Ag+(aq) 2Ag(s) + Cd2+(aq)

Formula of the species oxidised = Cd

State of the species: (s)

4. 3Zn(s) + 2Au3+(aq) + 6Cl-(aq) 2Au(s) + 3Zn2+(aq) + 6Cl-(aq)

In the equation, Zn is oxidised from 0 to +2.

Au is reduced from +3 to 0.

precursor donating electrons means oxidation.

Formula of the product formed by donating electrons: Zn2+

oxidation state of the product: (aq.)

5. Ba(s) + H2O(l) Ba2+(aq) + OH-(aq) + H2(g)

Ba is oxidised from 0 to +2.

H is reduced from +1 to 0.

The half reaction is: 2H2O + 2e- -------------> H2 + 2OH-

Reduction product is H2.

Element: H

Oxidation state: 0

6. 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

In the equation, Na is oxidised from 0 to +1.

H is reduced from +1 to 0.

In H2O, oxidation state of H is +1 and oxidation state of H in H2 is 0.

Symbol: H

Initial oxidation state: +1

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.