Question

Consider the reaction $C{r}_2{O}_{\mathit{7}}^{\mathit{2}\mathit{-}}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$. What is the quantity of electricity in coulombs needed to reduce $1$ mol of $C{r}_2{O}_{\mathit{7}}^{\mathit{2}\mathit{-}}$?

• A. $6\times {10}^6\mathrm{C}$
• B. $5.79\times {10}^5\mathrm{C}$
• C. $5.25\mathrm{x}{10}^5\mathrm{C}$
• D. None of these

Answer 1

The correct option is B

$5.79\times {10}^5\mathrm{C}$

The quantity of electricity can be calculated by calculating the number of mols electrons. As the flow of electrons is called electricity.

Step-1: Calculate mols of electrons needed.

• In the given reaction, it can be observed that 1 mol of ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ requires 6 mols of ${\mathrm{e}}^{-}$.

Step-2: Calculate the quantity of electricity:

• $1$ mol of electron contains $96500\mathrm{C}$ of charge.
• So, $6$ mols of ${\mathrm{e}}^{-}$ will have $6\times 96500=579000\mathrm{C}$
• The total charge on $6\mathrm{mol}$ of electron can be rewritten as: $579000\mathrm{C}=5.79\times {10}^5\mathrm{C}$

Hence, Option (B) is correct.