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B
Cl2+2I−→I2+2Cl−
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C
K++Na→K+Na+
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D
K+N→K−+N+
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Solution
The correct options are ACl2+2I−→I2+2Cl− B3Mg+2Al3+→2Al+3Mg2+
From the table, it can be seen that as Mg has a higher oxidising potential than Al, it will have a higher tendency to get oxidised. This means that it will act a reducing agent:
3Mg+2Al3+→2Al+3Mg2+
Similarly, chlorine has high reduction potential so it will oxidise iodide ion: