Question

How many mL of 0.125 M $C{{r}^{3+}}$ must be reacted with 12.00 mL of 0.200 M $MnO_{4}^{-}$ if the redox products are $C{{r}_{2}}O_{7}^{2-}$ and $M{{n}^{2+}}$?
(A) 8 mL
(B) 16 mL
(C) 24 mL
(D) 32 mL

Answer 1

The value of chromium ion can be calculated by equating the amount of chromium ion to the amount of permanganate ion. The amount can be calculated by multiplying the molarity into the volume of the substance into the equivalent.

Complete step by step solution:
So the question says, $C{{r}^{3+}}$ and $MnO_{4}^{-}$ combine to form $C{{r}_{2}}O_{7}^{2-}$and $M{{n}^{2+}}$. So, the balanced equation will be:
$C{{r}^{3+}}+MnO_{4}^{-}\to M{{n}^{2+}}+\dfrac{1}{2}C{{r}_{2}}O_{7}^{2-}$
So, to calculate the amount of chromium ion needed, we have to calculate the equivalent of chromium ion and permanganate ion.
The equivalent of chromium is equal to the change in the oxidation number of chromium ion multiplied to the number of moles.
The oxidation number of chromium ion ($C{{r}^{3+}}$) is:
$x=+3$
The oxidation number in $C{{r}_{2}}O_{7}^{2-}$is :
$2x+7(-2)=-2$
$x=+6$
So the change in oxidation number is equal to 3
$Equivalent\text{ }of\text{ }C{{r}^{3+}}=3\text{ x }moles\text{ }of\,C{{r}^{3+}}$
The number of moles of $C{{r}^{3+}}$= 1
Equivalent of $C{{r}^{3+}}$= 3
The equivalent of permanganate is equal to the change in oxidation number of manganese atoms multiplied to the number of moles.
The oxidation number of manganese in ($MnO_{4}^{-}$) is:
$x+4(-2)=-1$
$x=+7$
The oxidation number of manganese in $M{{n}^{2+}}$ is:
$x=+2$
So, change in oxidation state is equal to 5
$Equivalent\text{ }of\text{ }MnO_{4}^{-}=5\text{ x }moles\text{ }of\,MnO_{4}^{-}$
The number of moles of $MnO_{4}^{-}$= 1
Equivalent of $MnO_{4}^{-}$= 5
The amount of $C{{r}^{3+}}$ will be equal to the molarity multiplied to the volume and equivalent.
$Amount\ of\text{ }C{{r}^{3+}}\text{=0}\text{.125 x }V\text{ x }equivalent$
The amount of $MnO_{4}^{-}$ will be equal to the molarity multiplied to the volume and equivalent.
$Amount\ of\text{ }MnO_{4}^{-}\text{ = 0}\text{.200 x 12 x }equivalent$
Equation both, we get:
$\text{0}\text{.125 x }V\text{ x 3}=0.200\text{ x 12 x 5}$
$V=32$
So, the volume of $C{{r}^{3+}}$ is 32 mL.

The correct answer is (D)- 32 mL.

Note: You should take the value of change in the number of oxidation numbers, not the oxidation number of the substance to be calculated. Equivalent mass is the mass of the substance that will combine with others.