1.

Find out in which option the order of reducing power is correct.

Answer»

`Cr^(3+) lt CL^(-) lt Mn^(2+) lt Cr`
`Mn^(2+) lt Cl^(-) lt Cr^(3+) lt Cr`
`Cr^(3+) lt Cl^(-) lt Cr_(2)O_(7)^(2-) lt MnO_(4)^(-)`
`Mn^(2+) lt Cr^(3+) lt Cl^(-) lt Cr`

SOLUTION :Less is the OXIDATION potential, weaker is the reducing agent. Oxidation potentials are `Cr^(3+)//Cr_(2)O_(7)^(2-)=1.33V,Cl^(-)//Cl_(2)=-1.36V,Mn^(2+)//MnO_(4)^(-)=-1.51V,Cr//Cr^(3+)=+0.74V`. Hence, ORDEROF reducing power will be `Mn^(2+)ltCl^(-)ltCr^(3+)ltCr`.


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