Pt(s) | H2(g) | H+(aq) || SO42-(aq) | PbSO4(s) | Pb(s)
Which balanced chemical equation, for an overall reaction, corresponds to this cell notation?
Notation => [Oxidation Rxn||Reduction Rxn]
[Pt(s) | H₂(g) | H⁺(aq) || SO₄ˉ²(aq) | PbSO₄(s) | Pb(s)]
H₂(g) + Pt(s) => 2H⁺(aq) + 2eˉ (Oxidation Half Rxn)
PbSO₄(s) + 2eˉ => Pb⁰(s) + SO₄ˉ²(aq) (Reduction Half Rxn)
…………………………………………………………..………..
H₂(g) + PbSO₄(s) + Pt(s)* => Pb⁰(s) + SO₄ˉ²(aq) + 2H⁺(aq)
*Pt(s) => Catalyst
The given cell notation represents a galvanic cell. The balanced chemical equation for the overall reaction can be determined by looking at the changes occurring at the two electrodes.
At the anode (left side of the cell notation), the species undergoing oxidation is H2(g), which is converted to H+ ions. The half-equation representing this process is:
H2(g) -> 2H+(aq) + 2e-
At the cathode (right side of the cell notation), the species undergoing reduction is PbSO4(aq). It is converted to solid PbSO4(s), and then further reduced to Pb(s). The half-equations representing these processes are:
PbSO4(aq) + 2e- -> PbSO4(s)
PbSO4(s) + 2e- -> Pb(s)
Combining these half-equations, we can cancel out the electrons and write the balanced overall reaction:
H2(g) + PbSO4(aq) -> 2H+(aq) + PbSO4(s) + Pb(s)
Therefore, the balanced chemical equation corresponding to the given cell notation is:
H2(g) + PbSO4(aq) -> 2H+(aq) + PbSO4(s) + Pb(s)
To determine the balanced chemical equation corresponding to this cell notation, it's important to understand that the vertical lines (|) in the cell notation represent the phase changes in the reaction. The double vertical lines (||) separate the reactants from the products.
In this given cell notation:
Pt(s) | H2(g) | H+(aq) || SO42-(aq) | PbSO4(s) | Pb(s)
The left side of the double vertical lines represents the anode, which is where oxidation occurs, while the right side represents the cathode, where reduction takes place.
From this information, we can identify the following half-reactions:
At the anode (oxidation):
H2(g) → 2H+(aq) + 2e-
At the cathode (reduction):
PbSO4(s) + 2e- → Pb(s) + SO42-(aq)
Now that we have the half-reactions, we can combine them to form the overall balanced equation. To do this, we need to ensure that the electrons cancel out. Therefore, we will multiply the half-reactions as needed:
2H2(g) → 4H+(aq) + 4e- (multiplied by 2)
PbSO4(s) + 2e- → Pb(s) + SO42-(aq)
Now, we can add the two half-reactions together:
2H2(g) + PbSO4(s) → 4H+(aq) + Pb(s) + SO42-(aq)
This balanced equation represents the overall reaction corresponding to the given cell notation.