Question

Consider the reaction:
$C{{r}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O$
What is the quantity of electricity in Coulombs needed to reduce 1mole of $C{{r}_{2}}{{O}_{7}}^{2-}$?
A) $6\times {{10}^{6}}C$
B) $5.79\times {{10}^{5}}C$
C) $5.25\times {{10}^{5}}C$
D) None of these.

Answer 1

1 Faraday (F) is equal to 96500C, Coulomb is the unit of charge.
This reaction is a reduction reaction.

Complete Solution :
In the question, we asked to find the quantity of electricity in Coulombs needed to reduce 1 mole of $C{{r}_{2}}{{O}_{7}}^{2-}$, so from that data we know the Chromium is getting reduced in this reaction.

Now let’s consider the reduction reaction that involves Cr,
$C{{r}_{2}}{{O}_{7}}^{2-}\to 2C{{r}^{3+}}$
Find the oxidation number of Chromium in $C{{r}_{2}}{{O}_{7}}^{2-}$ and $C{{r}^{3+}}$
Here Cr is getting reduced from ${{6}^{+}}\to {{3}^{+}}$ .
$2C{{r}^{6+}}+6{{e}^{-}}\to 2C{{r}^{3+}}$

So hereby we know that 6 electrons are involved in the reduction of chromium.
We know I Faraday (F) = 95000C = 1 mole of ${{e}^{-}}$
1 mole of ${{e}^{-}}$ $\to$ 96500 C
1 mole of $C{{r}_{2}}{{O}_{7}}^{2-}$requires 6 mole of ${{e}^{-}}$ to get reduced to +3 state.
1 mole of $C{{r}_{2}}{{O}_{7}}^{2-}$ $\to$ 6 mole of ${{e}^{-}}$ is required
6 mole of ${{e}^{-}}$ $\to$ $6\times 96500C$ = 579000 C
=$5.79\times {{10}^{5}}C$
So, the correct answer is “Option B”.

Note: Another way of solving this problem is:
We know that the valency factor for Cr is 6, valency factor is simply the N-factor i.e. change in oxidation state of the element while undergoing a reaction.
By Applying Faraday’s first law,
$\dfrac{Q}{F}=mole\times V.f$
Where Q is the charge, F is faraday, V.f is the valency factor
$Q=mole\times V.f\times F$
$Q=1\times 6\times 96500$
$Q=5.79\times {{10}^{5}}C$