Question

Suppose a current of 250.mA is passed through an electroplating cell with an aqueous solution of Ag2SO4 in the cathode compartment for 52.0 seconds. Calculate the mass of pure silver deposited on a metal object made into the cathode of the cell.

Answer 1

Answer:

$14.53~mg~of~Ag$

Explanation:

The first step is to know what is the half-reaction of silver, so:

$Ag_2SO_4_(_a_q_)-> Ag^+_(_a_q_)~+~SO_4^-^2_(_a_q_)$

Therefore, the half-reaction is:

$Ag^+_(_a_q_)~+~1e^-~->~Ag_(_s_)$

The next step is the calculation of the amount of charge, so:

$250~mA~\frac{1~A}{1000~mA}*52~s=13~C$

Now, we can calculate the mass, we have to keep in mind that we have 1 mol of electrons for each mol of $Ag^+$

$13~C\frac{1~mol~e^-}{96500~C}\frac{1~mol~Ag}{1~mol~e-}\frac{107.87~g~of~Ag}{1~mol~Ag}\frac{1000~mg~Ag}{1~g~Ag}=14.53~mg~Ag$