A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

To solve this problem, we need to use Faraday's laws of electrolysis.

Step 1: Convert time to seconds 20 minutes = 20 * 60 = 1200 seconds

Step 2: Calculate the total charge passed The total charge (Q) passed is given by the formula Q = It, where I is the current and t is the time. So, Q = 5 amperes * 1200 seconds = 6000 Coulombs

Step 3: Calculate the number of moles of electrons transferred The number of moles of electrons (n) is given by the formula n = Q/F, where F is Faraday's constant (96485 C/mol). So, n = 6000 C / 96485 C/mol = 0.0622 mol

Step 4: Determine the number of moles of Ni deposited The reaction at the cathode is Ni2+ + 2e- -> Ni. This means that for every 2 moles of electrons, 1 mole of Ni is deposited. So, the number of moles of Ni deposited is 0.0622 mol / 2 = 0.0311 mol

Step 5: Calculate the mass of Ni deposited The mass (m) is given by the formula m = nM, where M is the molar mass of Ni (58.69 g/mol). So, m = 0.0311 mol * 58.69 g/mol = 1.82 g

Therefore, the mass of Ni deposited at the cathode is 1.82 g.