It requires 40ml of 0.5 M Ce4+ to titrate 10ml of 1M Sn2+ to Sn4+. The oxidation state of cerium in the reduced product is:
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Final Answer :
The reduced oxidation state of Cerium in the reduced product is Ce3+.
Understanding :
1) Law of equivalents.
n -factor of Sn2+ = 4-2 = 2
x - change in Oxidation state of Ce4+
No. of equivalents of Ce(4+) = No. of equivalents of Sn(2+)
=> 40 * 0.5 * x = 10 * 2 * 1
=> x = 1 .
Therefore, oxidation state of Ce4+ is reduced by 1 unit I. e Ce(3+) .
The reduced oxidation state of Cerium in the reduced product is Ce3+.
Understanding :
1) Law of equivalents.
n -factor of Sn2+ = 4-2 = 2
x - change in Oxidation state of Ce4+
No. of equivalents of Ce(4+) = No. of equivalents of Sn(2+)
=> 40 * 0.5 * x = 10 * 2 * 1
=> x = 1 .
Therefore, oxidation state of Ce4+ is reduced by 1 unit I. e Ce(3+) .
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