Solve this equation by oxidation-reduction method, balance this equati

Solve this equation by oxidation-reduction method, balance this equation basically. Cu plus HNO3 goes to Cu dinitrate, Cu bracket NO3 bracket closed twice, plus NO2 plus H2O. And then, solve the equation, balance the equation by half-cell reaction. The equation is Au plus 3I plus H2 gas goes to H plus I plus Au atom. And the other question for ion-electron balance, half-cell equation for balance is NO3 minus goes to NO, and the other half reaction is Br minus 1 ion goes to Br2, bromine gas.

Let me parse the three questions carefully before solving:

Q1 — Balance Cu + HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O by the oxidation number change method.
Q2 — Balance Au + I⁻ + H₂ → H⁺ + I⁻ + Au by the half-cell reaction method. (Note: the equation as stated — "Au + 3I⁻ + H₂ → H⁺ + I⁻ + Au" — doesn't make chemical sense as written; the most standard reaction fitting this description is the dissolution/formation of the gold-iodide complex, or it may be an AuI₃ reduction. I'll interpret it as the most chemically meaningful version and solve it.)
Q3 — Balance these two individual half-reactions by the ion-electron method:
- (a) NO₃⁻ → NO
- (b) Br⁻ → Br₂

Question 1 — Cu + HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O

Oxidation Number Change Method

Step 1 — Assign oxidation numbers

Species	Element	O.N.
Cu (metal)	Cu	0
HNO₃	N	+5
Cu(NO₃)₂	Cu	+2
NO₂	N	+4

Step 2 — Identify changes

Cu: 0 → +2 → loses 2 electrons (oxidation) ↑ by 2
N in HNO₃: +5 → +4 in NO₂ → gains 1 electron (reduction) ↓ by 1

Step 3 — Equalize electron transfer

To make electrons lost = electrons gained:

1 Cu atom loses 2 e⁻
2 N atoms (in 2 HNO₃) each gain 1 e⁻ → total 2 e⁻ gained

So the ratio is 1 Cu : 2 NO₂

Step 4 — Write with coefficients from electron balance

$$\mathbf{1},\text{Cu} + \text{HNO}_3 \rightarrow \mathbf{1},\text{Cu(NO}_3)_2 + \mathbf{2},\text{NO}_2 + \text{H}_2\text{O}$$

Step 5 — Balance the rest

Cu(NO₃)₂ needs 2 NO₃⁻ groups. The 2 NO₃⁻ in Cu(NO₃)₂ plus 2 HNO₃ that became NO₂ = 4 HNO₃ total on the left.
H balance: 4 HNO₃ → 4 H → 2 H₂O
O check: Left: 4 × 3 = 12 O. Right: Cu(NO₃)₂ has 6 O + NO₂ has 2 × 2 = 4 O + H₂O has 2 O = 12 O ✓

✅ Balanced Equation

$$\boxed{Cu + 4,HNO_3 \rightarrow Cu(NO_3)_2 + 2,NO_2 + 2,H_2O}$$

---## Question 2 — Half-Cell Reaction Method

Interpreting the equation as written: Au + I⁻ + H₂ → H⁺ + AuI (complex) + ...

The most chemically sensible reaction fitting "Au + I⁻ + H₂ → H⁺ + I⁻ + Au" is ambiguous. The most classic half-cell pair that makes sense here is:

Oxidation half: H₂ → H⁺ (H₂ is oxidized)

Reduction half: Au⁺ + e⁻ → Au (or AuI₃ + 3e⁻ → Au + 3I⁻)

The most likely intended reaction (common in electrochemistry textbooks) is:

Au³⁺ + 3I⁻ + (3/2)H₂ → Au + 3HI or equivalently:

2Au³⁺ + 6I⁻ + 3H₂ → 2Au + 6HI

I'll solve this as AuI₃ dissolves / Au³⁺ reduced by H₂ using the half-cell method.

Half-Cell Reaction Method

Identify oxidation states:

Au: Au³⁺ (as AuI₃ or Au³⁺ in solution) → Au⁰: reduction (gains 3 e⁻)
H₂: H₂ (0) → 2H⁺ (+1): oxidation (loses 2 e⁻)

Half-reaction 1 — Reduction (cathodic):

$$\text{Au}^{3+} + 3e^- \rightarrow \text{Au}^0$$

Half-reaction 2 — Oxidation (anodic):

$$\text{H}_2 \rightarrow 2\text{H}^+ + 2e^-$$

Equalize electrons (LCM of 3 and 2 = 6):

Multiply reduction half × 2: $$2\text{Au}^{3+} + 6e^- \rightarrow 2\text{Au}$$

Multiply oxidation half × 3: $$3\text{H}_2 \rightarrow 6\text{H}^+ + 6e^-$$

Add the two half-reactions (cancel 6e⁻):

$$2\text{Au}^{3+} + 3\text{H}_2 \rightarrow 2\text{Au} + 6\text{H}^+$$

Including the iodide (since Au³⁺ is presented with I⁻ as AuI₃ complex):

$$2\text{AuI}_3 + 3\text{H}_2 \rightarrow 2\text{Au} + 6\text{HI}$$

Or in ionic form:

✅ Balanced Equation

$$\boxed{2,\text{Au}^{3+} + 6,\text{I}^- + 3,\text{H}_2 \rightarrow 2,\text{Au} + 6,\text{H}^+ + 6,\text{I}^-}$$

Simplified (cancel 6I⁻ on both sides — they are spectators):

$$\boxed{2,\text{Au}^{3+} + 3,\text{H}_2 \rightarrow 2,\text{Au} + 6,\text{H}^+}$$

---## Question 3 — Ion-Electron (Half-Reaction) Balancing

Rules for ion-electron method in acidic solution:

Balance atoms other than O and H first.
Balance O by adding H₂O.
Balance H by adding H⁺.
Balance charge by adding e⁻.

Half-reaction (a): NO₃⁻ → NO (Reduction, acidic solution)

Step 1 — Balance N: Already 1 N each. ✓

Step 2 — Balance O: Left has 3 O, right has 1 O → add 2 H₂O to the right: $$\text{NO}_3^- \rightarrow \text{NO} + 2\text{H}_2\text{O}$$

Step 3 — Balance H: Right now has 4 H → add 4 H⁺ to the left: $$\text{NO}_3^- + 4\text{H}^+ \rightarrow \text{NO} + 2\text{H}_2\text{O}$$

Step 4 — Balance charge:

Left charge: (−1) + 4(+1) = +3
Right charge: 0 + 0 = 0
Difference = 3 → add 3e⁻ to the left:

$$\boxed{\text{NO}_3^- + 4\text{H}^+ + 3e^- \rightarrow \text{NO} + 2\text{H}_2\text{O}}$$

Verification:

Atoms: N✓, O: 3=3✓, H: 4=4✓
Charge: (−1+4−3)=0 on left; 0 on right ✓

Half-reaction (b): Br⁻ → Br₂ (Oxidation, acidic solution)

Step 1 — Balance Br: Right has 2 Br → put 2 Br⁻ on the left: $$2\text{Br}^- \rightarrow \text{Br}_2$$

Step 2 — Balance O: No oxygen present. ✓

Step 3 — Balance H: No hydrogen. ✓

Step 4 — Balance charge:

Left charge: 2(−1) = −2
Right charge: 0
Difference = 2 → add 2e⁻ to the right (since it's oxidation, electrons are a product):

$$\boxed{2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-}$$

Verification:

Atoms: Br: 2=2 ✓
Charge: −2 on left; 0+(−2) = −2 on right ✓

Summary of All Three Solutions

Q1 — Oxidation Number Change Method

Step	Detail
Cu: 0 → +2	Loses 2e⁻ (oxidized)
N: +5 → +4	Gains 1e⁻ (reduced)
Equalize	1 Cu : 2 N → ratio 1:2

$$\boxed{Cu + 4,HNO_3 \rightarrow Cu(NO_3)_2 + 2,NO_2\uparrow + 2,H_2O}$$

Q2 — Half-Cell Reaction Method (Au³⁺/H₂)

Half-Reaction	× factor
Au³⁺ + 3e⁻ → Au	× 2
H₂ → 2H⁺ + 2e⁻	× 3

$$\boxed{2,Au^{3+} + 3,H_2 \rightarrow 2,Au + 6,H^+}$$

Q3 — Ion-Electron Half-Reactions

(a) Reduction: $$\boxed{NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O}$$

(b) Oxidation: $$\boxed{2Br^- \rightarrow Br_2 + 2e^-}$$

Combined overall reaction (if these two half-reactions are paired — multiply (a)×2 and (b)×3 to cancel 6e⁻):

$$2,NO_3^- + 8H^+ + 6Br^- \rightarrow 2,NO + 4,H_2O + 3,Br_2$$

Give a final equation written of all three, and there is no I minus or I type of thing in the AU wale equation. There is just AU plus 3 as H2 goes to H plus plus AU. That's it.

Got it. The Au equation is simply:

Au + 3H₂ → H⁺ + Au — meaning Au is a catalyst/spectator and H₂ is oxidized to H⁺. But since Au appears unchanged on both sides, the net ionic reaction is just the hydrogen oxidation. Let me re-interpret this properly:

The equation is Au³⁺ + 3/2 H₂ → Au + 3H⁺, balanced by half-cell method as shown below.

✅ Final Balanced Equations — All Three

1️⃣ Cu + HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O

Method: Oxidation Number Change

Cu: 0 → +2 (loses 2e⁻)
N: +5 → +4 (gains 1e⁻)
Equalize: 1 Cu : 2 NO₂ → multiply accordingly, balance H and O with H₂O

$$\boxed{Cu + 4,HNO_3 \rightarrow Cu(NO_3)_2 + 2,NO_2 + 2,H_2O}$$

2️⃣ Au³⁺ + H₂ → H⁺ + Au

Method: Half-Cell Reaction

Reduction half (×2): $$2Au^{3+} + 6e^- \rightarrow 2Au$$

Oxidation half (×3): $$3H_2 \rightarrow 6H^+ + 6e^-$$

Add and cancel 6e⁻:

$$\boxed{2,Au^{3+} + 3,H_2 \rightarrow 2,Au + 6,H^+}$$

3️⃣ Half-Cell Equations by Ion-Electron Method

(a) NO₃⁻ → NO Balance O with H₂O, H with H⁺, then charge with e⁻:

$$\boxed{NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O}$$

(b) Br⁻ → Br₂ Balance Br atoms, then charge with e⁻:

$$\boxed{2Br^- \rightarrow Br_2 + 2e^-}$$

You have to add the equation of NO3 and Br minus 1. It will form one equation, the half-cell equation. So you have to complete one equation from these two. So sum up in one equation after canceling everything.

Combining the Two Half-Reactions into One Overall Equation

The Two Half-Reactions:

(Reduction): NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O

(Oxidation): 2Br⁻ → Br₂ + 2e⁻

Step 1 — Equalize electrons (LCM of 3 and 2 = 6):

Multiply reduction × 2: $$2NO_3^- + 8H^+ + 6e^- \rightarrow 2NO + 4H_2O$$

Multiply oxidation × 3: $$6Br^- \rightarrow 3Br_2 + 6e^-$$

Step 2 — Add both half-reactions and cancel 6e⁻:

$$2NO_3^- + 8H^+ + 6e^- + 6Br^- \rightarrow 2NO + 4H_2O + 3Br_2 + 6e^-$$

✅ Final Overall Balanced Equation:

$$\boxed{2NO_3^- + 8H^+ + 6Br^- \rightarrow 2NO + 3Br_2 + 4H_2O}$$

Verification:

	Left	Right
N	2	2 ✓
O	6	4 ✓ (6 from NO₃⁻ = 2 in NO + 4 in H₂O)
H	8	8 ✓
Br	6	6 ✓
Charge	(−2) + (+8) + (−6) = 0	0 + 0 + 0 = 0 ✓

Everything balances perfectly.

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