Question

An unknown metal forms a soluble compound, M(NO3)2, which then undergoes electrolysis with an applied current of 2.50 amperes. If the current is applied for 22.6 min, 1.87 g of metal is deposited. a. Is the metal deposited at the anode or cathode? b. Draw the reaction that occurs at the other electrode, where the metal is NOT being deposited. c. Calculate the molar mass of the metal M. What is the metal? d. Which halogen(s) could react with M to spontaneously oxidize it back to its 2+ cation?

Answer 1

The metal is the cathode

The metal is palladium

At the other cathode, the reaction going on is

M(s) ------> M2^+(aq) + 2e

Chlorine can spontaneously react with Pd converting it to Pd^2+

The molar mass of the metal is 106.4 g of M

Step-by-step explanation:

Given the formula of the compound, M^2+ ion is involved. The equation of the reaction is;

M^2+(aq) + 2e -------------> M(s)

The number of moles of electrons transferred = 2 moles

1 mole of electron = 96500C

And Q= It

Where I= current

t= time taken in seconds

Now;

x g of the metal (its molar mass) is deposited by 2×96500 C

1.87 g of the metal is deposited by (2.50× 22.6×60) C

x = 1.87 × 2×96500/ 2.50 × 22.6 × 60

x= 360910/3390

x= 106.4 g of M

The metal is the cathode

The metal is palladium

At the other cathode, the reaction going on is

M(s) ------> M2^+(aq) + 2e

Chlorine can spontaneously react with Pd converting it to Pd^2+