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[muttiah]

<TABLE bgColor=#ffff80 border=1><TBODY><TR><TD>Pb + SO<SUB>4</SUB><SUP>2-</SUP> —> PbSO<SUB>4</SUB> + 2e<SUP>- 1.</SUP></TD><TD> </TD></TR><TR><TD>PbO<SUB>2</SUB> + SO<SUB>4</SUB><SUP>2-</SUP> + 4H<SUP>+</SUP> + 2e<SUP>-</SUP> —> PbSO<SUB>4</SUB> + 2H<SUB>2</SUB>O</TD></TR></TBODY></TABLE>
how did they get these reactions for the anode and cathode?

 

[falcon07]

The lead-acid cell uses a sulphuric acid electrolyte, which is where the hydrogen and sulphate ions come from.

 

[muttiah]

how come there is sulpahate ions in both equations?

 

[falcon07]

Elemental lead has an oxidation number of 0. When it forms an ionic bond with a sulphate ion, its oxidation number increases to 2. This is the oxidation reaction that takes place at the anode. The cathode is lead (IV) oxide. The lead has an oxidation number of 4. When it forms an ionic bond with a sulphate ion, its oxidation number decreases to 2. This is the reduction reaction, which makes the net reaction a redox reaction even though sulphate ions are present in both half equations. The hydrogen ions from the acid react with the oxygens from the lead (IV) oxide to form water and their movement completes the circuit, allowing current to flow.