If a current of 1.5 A is passed for 4.00 hours through a molten tin salt and 13.3g of tin is deposited, what is the oxidation state of the metal in the salt?
(Sn = 118.7, F = 96500 C mol-1)
(Sn = 118.7, F = 96500 C mol-1)
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Correct Answer: Option B
Explanation:
Q = It = 1.5 x 4 x 60 x 60
13.3 g = 216000 C
118.7 g = (216000 x118.7) / (13.3) = 1927759.4 C
06500 C - 1 mole
1927759.4 C = (1927759.4) / (96500) = 1.998 = 2
Q = It = 1.5 x 4 x 60 x 60
13.3 g = 216000 C
118.7 g = (216000 x118.7) / (13.3) = 1927759.4 C
06500 C - 1 mole
1927759.4 C = (1927759.4) / (96500) = 1.998 = 2