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How to answer all these following questions?

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For redox reaction use the redox pair law
a) S₂O₃ 2- / S₄O₆ 2- and I₂/I-

2 S₂O₃ 2- => S₄O₆ 2-

Balance charge
2 S₂O₃ 2- => S₄O₆ 2- + 2e-

I₂ => 2 I-

Balance charge
I₂ + 2e- => 2 I-

Addition

2 S₂O₃ 2- + I₂=> S₄O₆ 2- + 2 I-
Try the other by yourself.

Answers to a), b) , C).

The final results are ok, I am wondering where the 2 H+ + 2e- comming from. Especially there is no acid in 1. and 2.

Fe 2+ => Fe 3+ + e- Oxidation
Ce 4+ +e - => Ce 3+ Reduction
Fe 2+ + Ce 4+ => Fe 3+ + Ce 3+ Addition
Sulfate is spectator ion
For completion multiply by 2 and add SO₄ 2-
2 FeSO₄ + 2 Ce(SO₄)₂ => Fe₂(SO₄)₃ + Ce₂(SO₄)₃

Fe 2+ => Fe 3+ + e- Oxidation

HNO₃ + 3 H+ + 3e- => 2 H₂O + NO Reduction

Multiplikation of Oxidation by 3
3 Fe 2+ + 3 Fe 3+ + 3 e-

3 Fe 2+ + HNO₃ + 3 H+ => 3 Fe 3+ + 2H₂O + NO Addition
Also hier sulfate is spectator but can be added accordingly

6 FeSO₄ + 2HNO₃ + 3 H₂SO₄ => 3 Fe₂(SO₄)₃ + 4 H₂O + 2 NO

Answers to d), e), f), g).

I dont understand how you develop the part equations. In e) for example you use OH- even the reaction takes place with HCl. With mathematics a lot can do even the final result is correct but generally its wrong.

To start like this

d) S 2- => S + 2 e- Oxidation
HNO₃ + 3 H+ +3 e- => NO + 2 H₂O Reduction

e) S 2- => S+ 2 e- Oxidation
Cr₂O₇ 2- + 6 HCl + 8 H+ + 6 e- => 2 Cr 3+ + 7 H₂O + 6 Cl- Reduction

f) Cl₂ + 12 OH- => 2 ClO₃- + 6 H₂O + 10 e- Oxidation
Cl₂ + 2 e- => 2 Cl- Reduction

g) Fe 2+ => Fe 3+ + e- Oxidation

O₂ 2- + 2H₂O + 2 e- => 4 OH- Reduction

In answer to e) , I used basic medium even though HCl is reactant. You said it is wrong. After using acidic medium, the answer is as follows: The balanced net ionic equations for h), i), j) are as follows:
h)\(Bi_2S_3 + 8HNO_3 \rightleftarrows 2Bi(NO_3)_3 + 2NO + 3S + 4H_2O\)

i)\(H_3AsO_4 + 4Zn + 8HNO_3 \rightleftarrows AsH_3 + 4Zn(NO_3)_2 + 4H_2O\)

j)\(2CrI_3 + 64NaOH + 27Cl_2 \rightleftarrows 2Na_2CrO_4 + 6NaIO_4 + 54NaCl + 32H_2O\)