We read your photo as:

Fill in the blanks with suitable verbs from the brackets. The army _____ the enemy camp at night. The river _____ across the town. Write the balanced cathode reaction for a lead-acid battery.

Chemistry

ScanSolve AI Answer

Step-by-step solution

5 steps

Here are the solutions for section G and the chemistry problem.

G. Fill in the blanks with suitable verbs from the brackets.

  1. The army attacked the enemy camp at night.
  2. The river flowed across the town.

Chemistry Problem

The reaction at the cathode of a lead-acid battery during discharge, starting with PbO2\text{PbO}_2 and consuming 2e−2e^-, to produce PbSO4\text{PbSO}_4 and H2O\text{H}_2\text{O} is:

  1. Identify the given components and the type of reaction.

    The problem shows PbO2\text{PbO}_2 at the cathode, 2e−2e^- consumed, and PbSO4+2H2O\text{PbSO}_4 + 2\text{H}_2\text{O} as products. This is a reduction half-reaction in a lead-acid battery.

  2. Write the unbalanced half-reaction with the given species.

    PbO2+2e−→PbSO4+2H2O\text{PbO}_2 + 2e^- \rightarrow \text{PbSO}_4 + 2\text{H}_2\text{O}
  3. Balance the atoms other than oxygen and hydrogen.

    Lead (Pb) is already balanced (1 on each side). Sulfur (S) is present in PbSO4\text{PbSO}_4 on the product side, so we need SO42−\text{SO}_4^{2-} on the reactant side.

    PbO2+SO42−+2e−→PbSO4+2H2O\text{PbO}_2 + \text{SO}_4^{2-} + 2e^- \rightarrow \text{PbSO}_4 + 2\text{H}_2\text{O}
  4. Balance oxygen atoms by adding water molecules (if needed) and hydrogen atoms by adding H+\text{H}^+ ions (since it's an acidic environment).

    Reactant side has 2 oxygen from PbO2\text{PbO}_2 and 4 oxygen from SO42−\text{SO}_4^{2-} (total 6 oxygen). Product side has 4 oxygen from PbSO4\text{PbSO}_4 and 2 oxygen from 2H2O2\text{H}_2\text{O} (total 6 oxygen). Oxygen is balanced.
    Product side has 4 hydrogen atoms from 2H2O2\text{H}_2\text{O}. Add 4H+4\text{H}^+ to the reactant side.

    PbO2(s)+SO42−(aq)+4H+(aq)+2e−→PbSO4(s)+2H2O(l)\text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4\text{H}^+(aq) + 2e^- \rightarrow \text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)
  5. Verify the charge balance.

    Reactant side charge: (0)+(−2)+(+4)+(−2)=0(0) + (-2) + (+4) + (-2) = 0
    Product side charge: (0)+(0)=0(0) + (0) = 0
    The charges are balanced.

    The complete balanced half-reaction is:

    PbO2(s)+SO42−(aq)+4H+(aq)+2e−→PbSO4(s)+2H2O(l)\boxed{\text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4\text{H}^+(aq) + 2e^- \rightarrow \text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)}
Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp