A current of 5.29 A is passed through a Sn(NO3)2 solution for 1.30 h . How much tin is plated out of the solution?

15.26 g of tin

Explanation:

Given that the equation of the process is;

Sn^2+(aq) + 2e > Sn(s)

Since 1e = 1F= 96500 C

but q =It

t is in hours so it must be converted to seconds, the SI unit of time

Hence;

119g of tin is deposited by 2 * 96500 C

xg of tin is deposited by 5.29 * 1.30 * 60 * 60

x = 119 * 5.29 * 1.30 * 60 * 60/ 2 * 96500

x = 2946106.8/193000

x= 15.26 g of tin

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