What mass of copper would be deposited by a current of 1. 0 ampere passing for 965 seconds through copper (ll) tetraoxosulphate (IV) solution? [Cu = 63.5; 1...

What mass of copper would be deposited by a current of 1. 0 ampere passing for 965 seconds through copper (ll) tetraoxosulphate (IV) solution? [Cu = 63.5; 1F = 96500C]

Answer Details

To solve this problem, we can use the formula: mass of substance deposited = (current × time × molar mass) ÷ (Faraday’s constant × number of electrons transferred) First, we need to find the number of moles of copper deposited by dividing the charge passed by the Faraday’s constant: Q = I × t = 1.0 A × 965 s = 965 C n = Q ÷ (Faraday’s constant × number of electrons transferred) n = 965 C ÷ (96500 C/mol × 2) n = 0.005 mol Next, we can use the number of moles to calculate the mass of copper using its molar mass: mass = n × molar mass mass = 0.005 mol × 63.5 g/mol mass = 0.318 g Therefore, the mass of copper deposited by the current is 0.318 g. The correct option is (a) 0.318g.