Question

The charge on silver ions is +1. Deduce the charge on the dichromate ion in $A{g}_{2}C{r}_2{O}_7$ . Write the ionic equation for the formation of silver dichromate in this reaction. State symbols are not required.

Answer 1

Hint: To answer this, remember that we can write ionic equations by balancing the number of net charges on both sides. Dichromate ion is a strong oxidising agent. Oxidising agents take up electrons and reduce themselves while oxidising the other substance.

Complete step by step answer:
Firstly, let us deduce the charge on the dichromate ion in $A{g}_{2}C{r}_2{O}_7$ and then we will proceed to write down the ionic equation.
In the question, it is given to us that charge on silver ions is +1. Here, we can see that we will have 2 silver ions. We have 7 oxygen atoms and each has a charge of -2. So, in total it will have a charge of -14.
 So, we can write that the charge on chromium ions will be- $2+2x-14=0$
So, solving this we will get that charge on each chromium ion is +6.
So, overall charge on the dichromate ion will be $+6\times 2-14$ i.e. -2.
So, the charge on dichromate ion in $A{g}_{2}C{r}_2{O}_7$ is -2.
Now, an ionic equation is similar to a molecular equation but here we write the molecules in their dissociated ion forms i.e. the form in which they exist in solutions. The net charge on both the sides is equal.
So let us write down the ionic equation for the formation of silver dichromate.
$$2A{g}^{+}+C{r}_2{O_7}^{2-}\to AgC{r}_2{O}_7$$
We can understand the required answer from the above discussion.

Note: Silver dichromate as well as potassium dichromate acts as a strong oxidising agent in acidic medium because it exists as dichromate ion in acidic medium. It has a negative charge of -2 therefore, it readily takes up electrons and oxidises other compounds. In the alkaline medium, it exists as chromate instead of dichromate.