Answer:
5.55g of Li will be deposited at the cathode.
Step-by-step explanation:
Step 1:
Data obtained from the question. This include:
Current (I) = 2.5A
Time (t) = 8.5hrs = 8.5 x 60 x 60 = 30600 secs
Step 2:
Determination of the quantity of electricity.
The quantity of electricity, Q can be obtained as follow:
Q = it
Q = 2.5 x 30600
Q = 76500C
Step 3:
Determination of the number of faraday needed to deposit metallic lithium.
Lithium is a univalent metal and will be deposited in solution as follow:
Li+ e- —> Li
From the above, 1 faraday is required to deposit metallic lithium.
1 faraday = 96500C
Molar mass of Li = 7g/mol
From the balanced equation above,
96500C of electricity is required to deposit 7g of Li.
Step 4:
Determination of the mass of lithium deposited during the process. This is illustrated below:
From the balanced equation above,
96500C of electricity is required to deposit 7g of Li.
Therefore, 76500C of electricity will deposit = (76500 x 7)/96500 = 5.55g of Li.
Therefore, 5.55g of Li will be deposited at the cathode.
Note: the cathode is the negative electrode and Li being a metal will migrate to the cathode while chlorine being a non metal will migrate to the positive electrode(anode).