Question

Dichromate ion in acidic medium oxidizes stannous ion as:
$xS{n^{ + 2}} + yC{r_2}{O_7}^{2 - } + z{H^ + } \to aS{n^{ + 4}} + bC{r^{ + 3}} + c{H_2}O$
A) The value of $x:y = 1:3$
B) The value of $x + y + z = 18$
C) $a:b = 3:2$
D) The value of $z - c = 7$

Answer 1

Hint: The given reaction is a redox reaction in which the Sn is getting Oxidized and Cr is getting reduced. The reaction is carried out in acidic medium, therefore we’ll first balance the redox reaction in acidic medium.

Complete answer:
The given reaction is a redox reaction. We’ll step by step balance the given redox reaction.
The oxidation no. of Sn in the reactant side is $ + 2$ and $ + 4$ on the product side. The $Cr$ is getting reduced from $ + 6 \to + 3$. The individual redox reactions can be given as:
Oxidation Reaction: $S{n^{ + 2}} \to S{n^{ + 4}} + 2{e^ - }$ -- (1)
Reduction Reaction: $C{r_2}{O_7}^{2 - } + 6{e^ - } \to 2C{r^{ + 3}}$ --(2)
Multiplying reaction (1) by 3 to balance the electrons we get,
$3S{n^{ + 2}} \to 3S{n^{ + 4}} + 6{e^ - }$ --(3)
The overall reaction can be given as:
$3S{n^{ + 2}} + C{r_2}{O_7}^{2 - } \to 3S{n^{ + 4}} + 2C{r^{ + 3}}$
Now, balancing the number of Oxygen atoms and Hydrogen atoms on both sides by adding water molecules and ${H^ + }$ ions. The overall reaction is:
$3S{n^{ + 2}} + C{r_2}{O_7}^{2 - } + 14{H^ + } \to 2C{r^{ + 3}} + 3S{n^{ + 4}} + 7{H_2}O$
The values of $x = 3,y = 1,z = 14,a = 3,b = 2,c = 7$
The value of $x:y = 3:1$, hence Option A is incorrect.
The value of $x + y + z = 3 + 1 + 14 = 18$. Hence Option B is correct.
The value of $a:b = 3:2$. Option C is correct.
The value of $z - c = 14 - 7 = 7$, Option D is correct.
And hence option B, C and D are correct.

Note:
This reaction gives a change in colour too, along with a change in oxidation number. The dichromate solution is initially yellow in colour. Tin (II) chloride or stannous chloride ($SnC{l_2}$) is a colourless solution. When this is added to a solution of dichromate the colour changes from yellow to green. Green colour is an indication of the formation of $C{r^{ + 3}}$. The green colour compound is Chromium Chloride ($CrC{l_3}$). The overall reaction can be given as:
${K_2}C{r_2}{O_7} + {H_2}S{O_4} + SnC{l_2} \to {H_2}O + {K_2}S{O_4} + CrC{l_3} + Sn{(S{O_4})_2}$