How do you complete this equation

5.55g of Li will be deposited at the cathode.

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).