Question

Consider the following 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 1 mole of $C{{r}_{2}}O_{7}^{2-}$?
A. $579000C$
B. $679000C$
C. $879000C$
D. $979000C$

Answer 1

Hint: Think about the oxidation numbers of chromium on the reactants side as well as the products side. Consider the conversion between Faradays and Coulombs and how they are related to the number of moles of electrons involved in the reaction.

Complete step by step solution:
Before moving onto the solution, let us first check the oxidation number of chromium when it is on the reactants side as well as the products side. From this, we will be able to figure out the number of moles of electrons that are involved in this reaction.
When chromium is on the reactants side of the reaction and bonded with oxygen, we know that the fixed oxidation number of oxygen is -2 and the net charge on the ion is -2. So, we can say that the oxidation state of each chromium atom is +6. On the products side, we can see clearly that it is +3. Thus, we have verified that chromium has been reduced and there are 6 moles of electrons involved in this reaction. Now, we know that the amount of electricity required to reduce 1 mole of $C{{r}_{2}}O_{7}^{2-}$ will be equivalent to the amount of charge present on the six moles of electrons. Faraday’s laws of electrolysis state that the amount of charge present on 1 mole of electrons is equal to 1 faraday, which is equal to $96500C$. Now to calculate the amount of electricity of charge required, we will multiply this value of a faraday by the number of moles of electrons involved in the reaction.
\[\begin{align}
& \text{electricity required}=6\times 1F \\
& \text{electricity required}=6\times 96500C \\
& \text{electricity required}=579000C \\
\end{align}\]

Hence, the correct answer to this question is ‘A. $579000C$’

Note: Remember that here, electricity and charge are equivalent terminologies. Electrons are the entities that make electricity flow and the charge present on these electrons give them their electrical properties. Hence, we can interchange the terms here.