Question

When $S{O_2}$ is passed through an acidified solution of potassium dichromate, then chromium sulphate is formed. The change in oxidation number of chromium is?
A) $+ 4$ to $+ 2$
B) $+ 5$ to $+ 3$
C) $+ 6$ to $+ 3$
D) $+ 7$ to $+ 2$

Answer 1

As we are well aware with the fact that potassium dichromate is a strong oxidising agent so it will oxidise the sulphur dioxide into potassium sulphate and chromium sulphate and we know that it is a redox reaction so oxidation and reduction will both take place in same reaction.

Complete step by step solution:
As we know that sulphur dioxide reacts with potassium dichromate in an acidified solution and results in the production of potassium sulphate, chromium sulphate and water molecules. We can show this reaction with the help of an equation:
$S{O_2} + {K_2}C{r_2}{O_7} + 3{H_2}S{O_4} \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + 3{H_2}O$

We can now calculate the oxidation states of these compounds as we know that oxygen has an oxidation state of $- 2$, hydrogen and potassium are present in $+ 1$ oxidation states so we can find the oxidation states of others.
Let the oxidation number of sulphur in sulphur dioxide be $x$ then,
$\Rightarrow x + ( - 2) \times 2 = 0$
$\Rightarrow x = + 4$
Similarly, the oxidation state of chromium in potassium dichromate would be:
$\Rightarrow + 2 + 2x + ( - 2) \times 7 = 0$
$\Rightarrow x = + 6$
So the chromium is present in $+ 6$oxidation state.
Now let us find the oxidation states of products. Let us assume the oxidation number of chromium as $x$ and we know that sulphate has an oxidation state of $- 2$, so chromium will have an oxidation state as calculated:
$\Rightarrow 2x + ( - 2) \times 3 = 0$
$\Rightarrow x = + 3$
Therefore from the above calculations we can see that the oxidation state of chromium is changed from $+ 6$ to $+ 3$.
Hence, the correct answer is option (C).

Note: Always remember that oxidation states are the degree of oxidations of an atom in a chemical compound. It can be zero, positive or negative. The oxidation number can be taken as charge on the ion and oxidation number or state of a free element is always zero, monatomic ions have an atomic state equal to its charge and hydrogen and oxygen usually have $+ 1$ and $- 2$ oxidation states.