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.

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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