In an electrolysis experiment, the same quantity of electricity deposited 27 g of silver (Ar: 108) and 14 g of cadmium (Ar: 112).
What is the charge on cadmium?
A) 2+
B) 3+
C) 4+
D) 5+
Answer: A
Note:
- same quantity of electricity = same no. of moles of electrons being supplied in both experiments
- Half-equations for the reduction of each metals ion at the cathode:
Ag+ + e– → Ag
Cdx+ + xe– → Cd
x is unknown.
No. of moles of Ag = 27/108 = 0.25 mol
Moles of e– = 0.25 mol
since 1 mol of e– will produce 1 mol of Ag.
No. of moles of Cd = 14/112 = 0.125 mol
Moles of e– = 0.25 mol(from above)
Since 0.25 mol of e– produced 0.125 mol of Cd, it takes 2 mol of e– to produce 1 mol of Cd i.e. Cd2+ + 2e– → Cd
x = 2.
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