Question

Acidified potassium manganate(VII) reacts with iron(II) ethanedioate, FeC204.
The reactions taking place are shown.
MnO4 + 8H+ + 5e + Mn2+ + 4H20
Fe2+ → Fe3+ + e
C202 → 2CO2 + 2e
How many moles of iron(II) ethanedioate react with one mole of potassium manganate(VII)?

A 0.60 B 1.67 C 2.50 D 5.00​

Answer 1

Answer : The correct option is, (B) 1.67 mole

Explanation :

(1) $MnO_4^-+8H^++5e^\rightarrow Mn^{2+}+4H_2O$

(2) $Fe^{2+}\rightarrow Fe^{3+}+1e^-$

(3) $C_2O_2\rightarrow 2CO_2+2e^-$

The reaction 1 is a reduction reaction and reaction 2 and 3 are oxidation reaction.

Now adding both oxidation reaction, we get:

$Fe^{2+}+C_2O_2\rightarrow Fe^{3+}+1e^-+2CO_2+2e^-$

$Fe^{2+}+C_2O_2\rightarrow Fe^{3+}+2CO_2+3e^-$

Thus,

Oxidation reaction : $Fe^{2+}+C_2O_2\rightarrow Fe^{3+}+2CO_2+3e^-$

Reduction reaction : $MnO_4^-+8H^++5e^\rightarrow Mn^{2+}+4H_2O$

In order to balance the chemical we are multiplying oxidation reaction by 5 and reduction reaction by 3 then added both the reaction, we get the balanced redox reaction.

Oxidation reaction : $5Fe^{2+}+5C_2O_2\rightarrow 5Fe^{3+}+10CO_2+15e^-$

Reduction reaction : $3MnO_4^-+24H^++15e^\rightarrow 3Mn^{2+}+12H_2O$

The balance redox reaction will be:

$5Fe^{2+}+5C_2O_2+3MnO_4^-+24H^+\rightarrow 5Fe^{3+}+10CO_2+3Mn^{2+}+12H_2O$

Now we have to calculate the number of moles of iron(II) ethanedioate react with one mole of potassium manganate(VII).

From the balanced chemical reaction, we conclude that

As, 3 moles of potassium manganate(VII) react with 5 moles of iron(II) ethanedioate

So, 1 mole of potassium manganate(VII) react with $\frac{5}{3}=1.67$ moles of iron(II) ethanedioate

Therefore, the number of moles of iron(II) ethanedioate are 1.67 moles.