Complete and balance each of the following half-reactions (steps 2-5 in half reaction method):
(a) cr2+ --> cr3+
(b) hg + br- --> hgbr4 2-
(c) zns --> zn + s2-
(d) h2 --> h2o (in basic solution)
(e) h2 --> h3o (in acidic solution)
(f) no3- --> hno2 (in acidic solution)
(g) mno2 --> mno4- (in basic solution)
(h) cl- --> clo3- (in acidic solution)

(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!

the answer would be c

the potential energy of the reactants is 200j.

from the energy diagram, the energy of the product formed is 350j; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.