this is a repost.
Balance the following ionic equation for acidic conditions. Identify the oxidizing agent and the reducing agent. Show the steps.
Hg(ℓ) + NO3^ (–1) (aq) + Cl^(–1) (aq) → HgCl4^(–2) (s) + NO2(g)
Hg(ℓ) + NO3^ (–1) (aq) + Cl^(–1) (aq) → HgCl4^(–2) (s) + NO2(g)
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NO3^- + 1e + 2H^+ ==> NO2 + H2O
Hg + 4Cl^- ==> HgCl4^2- + 2e
I have separated into the two half equations and I've balanced each.
Multiply equation 1 by 2 and add to equation 2 for the complete equation.
Cancel the two electrons on each side for the final equation.
I guess I'm 223 but you got an answer.
To balance the given ionic equation for acidic conditions, follow these steps:
Step 1: Write the unbalanced equation.
The unbalanced equation is:
Hg(ℓ) + NO3^– (aq) + Cl^– (aq) → HgCl4^2– (s) + NO2(g)
Step 2: Balance the atoms other than hydrogen and oxygen.
Looking at the equation, we see that we have one mercury (Hg) atom on the left side and one on the right side, so it is already balanced.
Next, let's balance the chlorine (Cl) atoms. There is one Cl atom on the left side and four on the right side in the HgCl4^2– compound. To balance the Cl atoms, we can place a coefficient of four (4) in front of HgCl4^2–. The equation becomes:
Hg(ℓ) + NO3^– (aq) + Cl^– (aq) → 4HgCl4^2– (s) + NO2(g)
Step 3: Balance the charges.
The equation now has a charge balance issue. On the left side, the total charge is zero. On the right side, the total charge is –8 (four Cl ions with a charge of –1 each).
To balance the charges, we need to introduce H+ ions as hydrogen ions (protons) to the left side of the equation. We will introduce H+ ions equal to the number of Cl ions on the right side, which in this case is four (4).
Hg(ℓ) + NO3^– (aq) + Cl^– (aq) + 4H+ (aq) → 4HgCl4^2– (s) + NO2(g)
Step 4: Balance the hydrogen (H) atoms.
The left side now has four hydrogen atoms from the H+ ions, while the right side has no hydrogen atoms. To balance the hydrogen atoms, we introduce four (4) H2O molecules on the right side.
Hg(ℓ) + NO3^– (aq) + Cl^– (aq) + 4H+ (aq) → 4HgCl4^2– (s) + NO2(g) + 4H2O(l)
Step 5: Balance the oxygen (O) atoms.
The left side now has three oxygen (O) atoms from the NO3^– ion, while the right side has nine oxygen atoms (four from HgCl4^2– and five from H2O). To balance the oxygen atoms, we introduce five (5) H2O molecules on the left side.
Hg(ℓ) + 5H2O(l) + NO3^– (aq) + Cl^– (aq) + 4H+ (aq) → 4HgCl4^2– (s) + NO2(g) + 4H2O(l)
Step 6: Combine like terms and simplify.
After combining like terms, the balanced equation for the given ionic equation under acidic conditions is:
Hg(ℓ) + 5H2O(l) + NO3^– (aq) + Cl^– (aq) + 4H+ (aq) → 4HgCl4^2– (s) + NO2(g) + 4H2O(l)
Now that we have balanced the equation, we can identify the oxidizing agent and the reducing agent.
The oxidizing agent is the species being reduced, meaning it is losing electrons. In this case, NO3^– (nitrate ion) is being reduced to NO2(g), therefore NO3^– is the oxidizing agent.
The reducing agent is the species being oxidized, meaning it is losing electrons. In this case, Hg(ℓ) is losing electrons to form HgCl4^2– (s), therefore Hg(ℓ) is the reducing agent.