Calculate the expected amount of Ni deposited on a zinc electrode under conditions of 2.00 V and a current of 5.00 amperes for 15.00 minutes.
Charge Passed, Q = I * t = 5.00 * 15.00 * 60 = 4500 C
The balanced reaction will be
Ni(2+)(aq) + Zn(s) ------------- Zn(2+)(aq) + Ni(s)
Since the change in oxidation state of Ni is 2, going from +2 to 0, we need 2F or 2 * 96500C of charge to deposit one mole of Nickel
Number of moles of Ni deposited = (4500)/(2*96500) = 0.02332 moles
Molar mass of Ni = 58.7 g/mol
Mass of Ni deposited = (Number of moles) * (Molar mass) = 0.02332 mol * 58.7 g/mol = 1.37 grams
Note - Post any doubts/queries in comments section.
Calculate the expected amount of Ni deposited on a zinc electrode under conditions of 2.00 V...