The time required (in hours) to reduce $3 \mathrm{~mol} \mathrm{of}$ $\mathrm{Fe}^{3+}$ ions to…

The time required (in hours) to reduce $3 \mathrm{~mol} \mathrm{of}$ $\mathrm{Fe}^{3+}$ ions to $\mathrm{Fe}^{2+}$ ions with 2.0 amperes of current is $\left(1 \mathrm{~F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right)$
  1. 30.2
  2. 40.2
  3. 10.2
  4. 15.2

Solution

$\mathrm{i}=2.0 \mathrm{~A}, \mathrm{t}=?$ $3 \mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow 3 \mathrm{Fe}^{2+}$ Charge of 3 moles of $\mathrm{e}^{-}=3 \times 96500=289,500 \mathrm{C}$. $\begin{aligned} & \Rightarrow \mathrm{t}=\frac{\mathrm{Q}}{\mathrm{i}}=\frac{289,500}{2.0}=144,750 \text { seconds } \\ & =40.208 \text { hours. }\end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 1)

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