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).

plasma is the state of matter where the substance is ionized and has no definite shape or volume. the fact of being ionized makes the substance a good electrical conductor.