Answer:
(a) Cr ²⁺ → Cr³⁺ + e⁻
(b) Hg + 4 Br⁻ → HgBr₄²⁻ + 2e⁻
(c) ZnS + 2e⁻ → Zn + S²⁻
(d) 2OH⁻ + H₂ → 2 H₂O + 2e ⁻(in basic solution)
(e) H₂ → 2 H⁺ + 2e⁻ (in acidic solution)
(f) 2e⁻ + 3H⁺ + NO₃⁻ → HNO₂ + H₂O (in acidic solution)
(g) 4OH⁻ + MnO₂ → MnO₄⁻ + 2H₂O + 3e⁻ (in basic solution)
(h) 3H₂O + Cl⁻ → ClO₃⁻ + 6H⁺ + 6e⁻ (in acidic solution)
Explanation:
(a) Cr ²⁺ → Cr³⁺
You should see than Cr the only unbalanced stuff is charge than you must balance with electrons:
Cr ²⁺ → Cr³⁺ + e⁻
(b) Hg + Br⁻ → HgBr₄²⁻
Here you have to balance in the first the bromine (4 bromines in each side). In the last you must balance charges with electrons like in (a) exercise:
Hg + 4 Br⁻ → HgBr₄²⁻ + 2e⁻
(c) ZnS → Zn + S²⁻
Here, you must balance just charges with electrons (Zn and S atoms are balanced:
ZnS + 2e⁻ → Zn + S²⁻
(d) H₂ → H₂O (in basic solution)
First, you should balance oxygen atoms with water:
H₂O+ H₂ → H₂O
Then, hydrogen with H⁺:
H₂O + H₂ → H₂O + 2H⁺
In basic solution you should add as many OH⁻ as H⁺ you have:
2OH⁻ + H₂O + H₂ → H₂O + (2H⁺ + 2OH⁻)
H⁺ + OH⁻ → H₂O. So:
2OH⁻ + H₂O + H₂ → H₂O + 2H₂O
Delete waters that are in both sides:
2OH⁻ + H₂ → 2H₂O
Last, balance charge with electrons
2OH⁻ + H₂ → 2 H₂O + 2e ⁻
(e) H₂ → H₂O (in acidic solution) -H₃O doesn't exist so I think is H₂O-
Balance oxygen with water
H₂O + H₂ → H₂O
Balance hydrogens with H⁺:
H₂O + H₂ → H₂O + 2H⁺ -you don't put OH⁻ because this is just for basic solution-
Delete waters:
H₂ → 2H⁺
Balance charge with electrons:
H₂ → 2 H⁺ + 2e⁻ (in acidic solution)
(f) NO₃⁻ → HNO₂ (in acidic solution)
Balance oxygen with water:
NO₃⁻ → HNO₂ + H₂O
Hydrogens with H⁺:
3H⁺ + NO₃⁻ → HNO₂ + H₂O
Charge with electrons:
2e⁻ + 3H⁺ + NO₃⁻ → HNO₂ + H₂O
(g) MnO₂ → MnO₄⁻ (in basic solution)
Balance oxygen with water:
2H₂O + MnO₂ → MnO₄⁻
Hydrogen with H⁺:
2H₂O + MnO₂ → MnO₄⁻ + 4H⁺
In basic solution you should add as many OH⁻ as H⁺ you have:
4OH⁻ + 2H₂O + MnO₂ → MnO₄⁻ + (4H⁺ + 4OH⁻ → 4H₂O)
Delete waters in both sides:
4OH⁻ + MnO₂ → MnO₄⁻ + 2H₂O
Balance charge with electrons:
4OH⁻ + MnO₂ → MnO₄⁻ + 2H₂O + 3e⁻
(h) Cl⁻ → ClO₃⁻ (in acidic solution)
Balance oxygen with waters:
3H₂O + Cl⁻ → ClO₃⁻
Hydrogens with H⁺:
3H₂O + Cl⁻ → ClO₃⁻ + 6H⁺
Charge with electrons:
3H₂O + Cl⁻ → ClO₃⁻ + 6H⁺ + 6e⁻
I hope it helps!
