When lead metal (Pb) is added to a solution containing chromium ions (\( \text{Cr}^{3+} \)), a redox reaction can occur because lead can act as a reducing agent. In this context, the lead metal will reduce the \( \text{Cr}^{3+} \) ions to chromium metal (\( \text{Cr} \)), while the lead itself will be oxidized to lead ions (\( \text{Pb}^{2+} \)).
The half-reactions for this process are as follows:
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Oxidation half-reaction (lead is oxidized): \[ \text{Pb} \rightarrow \text{Pb}^{2+} + 2\text{e}^- \]
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Reduction half-reaction (chromium is reduced): \[ \text{Cr}^{3+} + 3\text{e}^- \rightarrow \text{Cr} \]
To balance the overall reaction, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to have a balanced number of electrons exchanged:
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Oxidation (multiplied by 3): \[ 3\text{Pb} \rightarrow 3\text{Pb}^{2+} + 6\text{e}^- \]
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Reduction (multiplied by 2): \[ 2\text{Cr}^{3+} + 6\text{e}^- \rightarrow 2\text{Cr} \]
Now, we can combine these balanced half-reactions: \[ 3\text{Pb} + 2\text{Cr}^{3+} \rightarrow 3\text{Pb}^{2+} + 2\text{Cr} \]
This indicates that if enough lead is added to the chromium ions in the solution, the lead will reduce some of the \( \text{Cr}^{3+} \) ions to chromium metal, while it will itself be oxidized to \( \text{Pb}^{2+} \) ions.
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