Balance chemical reaction

[dashamshn]

Sn + NaNO₃ -> SnO₂ + NaN02
Equalize the reactions using the electron balance method, label the oxidizing agent and reducing agent

[ChenBeier]

What are your ideas.

Final is

Sn + 2 NaNO₃ => SnO₂ + 2 NaNO₂

[ChenBeier]

Divide by 2 is the result already given.

[michaeljordan]

Sn + 2 NaNO 3 => SnO 2 + 2 NaNO 2 is the most accurate.

[ChenBeier]

Why repetition. It's already written above.

[jebeloy]

copper(1) sulfate + sodium chloride

[ChenBeier]

Copper(I)sulfate is not existing.
The equilibrium is on left right side
Cu₂SO₄ =======> CuSO₄ + Cu

[saun]

To balance the given chemical reaction using the electron balance method, we first need to identify the oxidation states of the elements involved and determine which species are being oxidized and which are being reduced. The reaction is:
Sn+NaNO₃→SnO₂+NaNO₂

Step 1: Assign oxidation states

Sn: In its elemental form, Sn has an oxidation state of 0. In SnO₂SnO₂​, Sn has an oxidation state of +4.

Na: In NaNO₃NaNO₃​ and NaNO₂NaNO₂​, Na has an oxidation state of +1.

N: In NaNO₃NaNO₃​, N has an oxidation state of +5. In NaNO₂NaNO₂​, N has an oxidation state of +3.

O: Oxygen has an oxidation state of -2 in both compounds.

Step 2: Identify the oxidizing and reducing agents

Sn is being oxidized from 0 to +4 (loses 4 electrons).

N in NaNO₃NaNO₃​ is being reduced from +5 to +3 (gains 2 electrons).

Thus:

Oxidizing agent: NaNO₃NaNO₃​ (causes oxidation by accepting electrons).

Reducing agent: SnSn (causes reduction by donating electrons).

Step 3: Balance the electrons

Sn loses 4 electrons: Sn→Sn₄++4e−Sn→Sn₄++4e−.

Each N in NaNO₃NaNO₃​ gains 2 electrons: N₅++2e−→N₃+N₅++2e−→N₃+.

To balance the electrons, we need 2 N atoms to accept 4 electrons (since 2 × 2 = 4). Therefore, we need 2 NaNO₃NaNO₃​ molecules.

Step 4: Write the balanced equation

Sn+2NaNO₃→SnO₂+2NaNO₂
​
Step 5: Verify the balance

Sn: 1 on both sides.

Na: 2 on both sides.

N: 2 on both sides.

O: 6 on both sides.

The equation is balanced.

Final Answer:
Sn+2NaNO₃→SnO₂+2NaNO₂

Oxidizing agent: NaNO₃NaNO₃​

Reducing agent: SnSn

[ChenBeier]

The answer was already given above. No need to repeat it.