Question

$Mn{{O}_{2}}$ is oxidized anodically. The number of hours required to oxidize 26.08 g of $Mn{{O}_{2}}$ by a current of 0.5 A is:
[Atomic mass of Mn = 55]

Answer 1

In electrochemistry we can’t always measure the amount of flow of charges. Instead the concept of current is used. Current is the amount of electrical charge that flows per unit time. It is measured in Amperes (A).

Complete step by step solution:

For this we use Faraday's Constant, F. Faraday's Constant is given by the symbol F and is defined as the charge in coulombs (C) of 1 mole of electrons. Faraday's constant is approximately 96485 C mol-1. You can calculate F by multiplying the charge on one electron (1.602 x 10-19) by Avogadro's number (6.022 x 1023),
We have been given in the problem that the:
Mass of $Mn{{O}_{2}}$ is 26.08g.
Atomic mass of Mn is 55 amu.
Current = 0.5 A
$M{{n}^{4+}}+4{{e}^{-}}\to Mn$
We can see from the above that 4 mol of electrons give 1 mol of Magnesium, Mn.
Now we will put the numbers in. 1 mol of electrons is 1 faraday.
4 x 96500 coulombs give 55 g of copper.
Number of coulombs = $=\dfrac{26.08\times 2\times 96485}{55}=94502.86$C
Now we know how many coulombs we need, and the current was already given in the problem. Finally, we can calculate the time required to oxidize $Mn{{O}_{2}}$.
Number of coulombs = current in amps x time in seconds

& 94502.86=0.5\times \,t \\\ & t=18901.72\,\sec \\\ \end{aligned}$$ To convert into hours, we will divide by 60. $$\dfrac{18901.72}{60}=315\,hours$$ So, the total time taken will be 315 hours. **Note:** The same question can be given with time and asking to calculate the amount of $$Mn{{O}_{2}}$$ produced after oxidation. In that case the procedure should be: $$\text{Current}\,\text{and time}\to \text{Charge(C)}\to \text{faradays}\to \text{moles}\to \text{grams}$$