What current (in milliamperes, nearest integer) is required to deposit 0.195 g of platinum metal in 5 hours from a solution of PtCl6^(2-) ? (Atomic weight of Pt = 195)

1 mA
2 mA
3 mA
4 mA

Correct answer: 1 mA

Solution

From PtCl6^(2-), Pt is in the +4 state, requiring 4 electrons per atom. Applying Faraday's second law with the given values yields a current of approximately 1 mA.

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