Question

A metal bromide, MBr_2, is converted to a molten form at high temperature. Electrolysis of this sample with a current of 6.13A for 73.1 seconds results in deposition of 0.148 g of metal M at the cathode. The other product is bromine gas, Br_2(g), which is released at the anode. Determine the identity of metal M. Enter symbol of the element into clicker.

Answer 1

Answer:

M is copper (Cu)

Explanation:

Step 1:

Data obtained from the question. This includes:

Compound => MBr2

Current (I) = 6.13A

Time (t) = 73.1 secs

Mass of M = 0.148 g

Identity of M =?

Step 2:

Determination of the quantity of electricity that deposit 0.148 g of the metal M. This is illustrated below:

Current (I) = 6.13A

Time (t) = 73.1 secs

Quantity of electricity (Q) =?

Q= it

Q = 6.13 x 73.1

Q = 448.103C

Step 3:

Determination of the number of faraday and the quantity of electricity required to deposit the metal M. This is illustrated below:

In solution, the compound MBr2 will dissociate as follows

MBr2 —> M^2+ + 2Br^-

The metal M, will be deposited in the cathode according to the equation:

M^2+ + 2e- —> M

From the above, we can see clearly that 2 faraday is needed to deposit the metal M. Converting the number of faraday to electricity, we have:

1 faraday = 96500C

Therefore, 2 faraday = 2 x 96500C = 193000C

From the calculations made above, 193000C of electricity is needed to deposit the metal M.

Step 4:

Determination of the identity of the metal M. This is illustrated below:

From the calculations made above,

448.103C deposit 0.148 g of the metal M.

Therefore 193000C will deposit = (193000x0.148)/448.103 = 63.74g of the metal M.

Comparing the molar mass of M with the molar masses in the periodic table, M is copper (Cu).