Question

Some 1L electrolysis cells have a Fe3+ solution with a concentration of 0.0250M Mn2+. It is made up of Mn and Pt electrodes and the next electrochemical reaction occurs in the cell.

Mn (s) -> Mn2+ + 2e-

Fe3+ + 3e- -> Fe (s)

(1) Is Mn electrode anode or cathode

(2) If a current of 2.60A is applied to this cell for 18 minutes and 0.504g of Fe metal is released on the surface of the Pt electrode, calculate the total concentration of the Mn after the reaction.

Answer 1

1. Anode electrode means where oxidation occurs, Here Mn is getting oxidised so, Mn electrode is anode.

2. We know , no. of moles of a chemical species={given mass/molar mass}

Molar mass of Fe = 56 g/mol

So, 0.504 g Fe = {0.504/56}mol Fe = 0.009 mol Fe

For one mol Fe preparation we need 3 mol electron. Fe^{​​​​​3+} + 3 e^{​​​​​​-} = Fe

Hence, 0.009 mol Fe preparation will need (3×0.009) mol electron = 0.027 mol electron.

In the process electron is giving Mn and electron accepting Fe.

0.027 mol electron accepted by the Fe means 0.027 mol electron is given by Mn.

Now, Mn = Mn^{​​​​​2+} + 2 e^{​​​​​​-}

2 mol electron is given by 1 mol Mn

So, 0.027 mol electron is given by (1/2)×0.027 mol Mn = 0.0135 mol Mn.

And 1 mol Mn is converting to 1 mol Mn^{​​​​​2+}

So, 0.0135 mol Mn will convert to 0.0135 mol Mn^{​​​​​2+}.

Initial concentration of Mn2+ was 0.025 M

Now concentration of Mn2+ will be = (0.025+0.0135)M = 0.0385 M.

So total concentration of Mn2+ will be 0.0385M.