Question

Complete the following chemical equation:
$C{{r}_{2}}{{O}_{7}}^{2-}+6F{{e}^{2+}}+14{{H}^{+}}\to$

Answer 1

The dichromate ion has the chemical formula of $C{{r}_{2}}{{O}_{7}}^{2-}$. The dichromate ion is considered as the strong oxidising agent. In the acidic solution of the dichromate it tends to form the deep blue colour solution with the peroxide. It also has the tendency to react with the hydrogen sulphide and oxidise it to sulphur. It also oxidises the sulphites to the sulphates and chloride to the chlorine.

Complete step-by-step answer: As in the reaction we see that the hydrogen ions are present in the reaction that means the solution of dichromate is acidic. The dichromate has the tendency to oxidise the ion present in the solution. So here we have the ferrous ion in the solution which will be oxidised. Here we will use the half equation method so that after balancing both the half reactions on further adding them we get our equation. So first we will take the reduction half reaction that is the dichromate ion which will be reduced here and the hydrogen ion. so the reaction of reduction is as follows:
$C{{r}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}\to C{{r}^{3+}}+{{H}_{2}}O(l)$
Now firstly we will balance the equation:
$C{{r}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}\to 2C{{r}^{3+}}+7{{H}_{2}}O(l)$
Now we will check the charge whether it is balanced both on reactant and product side:
On the reactant side we have the charge equal to +12. Whereas on the product side we have the charge equal to +6. So we have to add 6 electrons on the reactant side. So the equation will be:
$C{{r}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O(l)$
Now we will oxidise the ferrous ion. so the oxidation half reaction is:
$F{{e}^{2+}}\to F{{e}^{3+}}$
Now the equation is balanced but the charge is not balanced as we have +2 charge on the reactant side and +3 charge on the product side. So to balance the charge we need to add an electron on the product side. So the equation will be:
$F{{e}^{2+}}\to F{{e}^{3+}}+1{{e}^{-}}$
So now we need that the electrons should be the same in both reduction and oxidation reactions so we will multiply the oxidation reaction by 6. We get:
$6F{{e}^{2+}}\to 6F{{e}^{3+}}+6{{e}^{-}}$
Now let us add both the reduction and oxidation equations we get:
$C{{r}_{2}}{{O}_{7}}^{2-}+6F{{e}^{2+}}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O(l)+6F{{e}^{3+}}+6{{e}^{-}}$
The electrons will cancel out each other and we will get our final equation as :
$C{{r}_{2}}{{O}_{7}}^{2-}+6F{{e}^{2+}}+14{{H}^{+}}\to 2C{{r}^{3+}}+6F{{e}^{3+}}+7{{H}_{2}}O(l)$
So the above equation is the final equation.

Note: The dichromate ion has the oxidation state equal to +6. It has the valency of 2. It is odourless substance which has the red orange crystalline appearance. It tends to get soluble in cold water. It helps in the leather industry for chrome tanning. It is used in volumetric estimation of the ferrous salts. It is also used in manufacturing of the chromium compounds.