1.) Cu+( a q )+IO( a q )Cu2+( a q )+I( a q ) Write the balanced half-reactions E

Question

1.) Cu+(aq)+IO(aq)Cu2+(aq)+I(aq)

Write the balanced half-reactions

Express your answer as a chemical equations separated by a comma. Identify all of the phases in your answer.

2.)Write a balanced redox equation

Express your answer as a chemical equation. Identify all of the phases in your answer.

3.) S2O23(aq)+Br2(l)S4O26(aq)+Br(aq)

Write the balanced half-reactions

Express your answer as a chemical equations separated by a comma. Identify all of the phases in your answer.

4.) Write a balanced redox equation

Express your answer as a chemical equation. Identify all of the phases in your answer.

5.) Mg(s)+VO34(aq)Mg2+(aq)+V2+(aq)

Write the balanced half-reactions

Express your answer as a chemical equations separated by a comma. Identify all of the phases in your answer.

6.) Write a balanced redox equation

Express your answer as a chemical equation. Identify all of the phases in your answer.

7.) Al(s)+Cr2O27(aq)Al3+(aq)+Cr3+(aq)

Write the balanced half-reactions

Express your answer as a chemical equations separated by a comma. Identify all of the phases in your answer.

8.) Write a balanced redox equation

Express your answer as a chemical equation. Identify all of the phases in your answer.

Explanation / Answer

1&2. Cu+(aq)+IO(aq)Cu2+(aq)+I(aq)

The oxidation state per atom is

Cu+(aq) is +1

IO(aq)is +1 (in which I has +1 and O has -2)

Cu2+(aq) is +2

I(aq) is -1

so we have

Oxidation: Cu+(aq)Cu2+(aq) + e-

Reduction: IO(aq) + 2e- I(aq)

To balance the reactions,

2*oxidation: 2Cu+(aq)2Cu2+(aq) + 2e-

Reduction: IO(aq) + 2e- I(aq)

Combining the two reactions

2Cu+(aq) + IO(aq) + 2e- 2Cu2+(aq) + I(aq) +2e-

after simplifying, fully balanced net ionic equation is

2Cu+(aq) + IO(aq) 2Cu2+(aq) + I(aq)

3&4. S2O23(aq)+Br2(l)S4O26(aq)+Br(aq)

The oxidation state per atom is

S2O23(aq) is -3 (in which S has +2 and O has -2)

Br2(l) is 0

S4O26(aq) is -6 (in which S is -2 and O is +2)

Br(aq) is -1

so we have

Oxidation: S2O23(aq) S4O26(aq) + 3e-

Reduction: Br2(l) + 2e- 2Br(aq)

To balance the reactions,

2*oxidation: 2S2O23(aq) 2S4O26(aq) + 6e-

3*reduction: 3Br2(l) + 6e- 6Br(aq)

Combining the two reactions

2S2O23(aq) + 3Br2(l) + 6e- 2S4O26(aq) + 6Br(aq) + 6e-

after simplifying, fully balanced net ionic equation is

2S2O23(aq) + 3Br2(l) 2S4O26(aq) + 6Br(aq)

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

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