Animated Solution for Chemistry - Electrochemistry: In an electrochemical cell, dichromate ions in aqueous acidic medium are reduced to ${\text{Cr}}^{3+}$. The current (in amperes) that flows through the cell for $48.25$ minutes to produce $1$ mole of ${\text{Cr}}^{3+}$ is _______. Use: $1$ Faraday = $96500{\text{ C mol}}^{-1}$

Imagine an electrochemical cell buzzing with energy. At the cathode, a fascinating transformation is taking place: vibrant orange dichromate ions (${\text{Cr}}_2{\text{O}}_7^{2-}$) are being systematically reduced to green chromium ions (${\text{Cr}}^{3+}$). This isn't just magic; it's a precise dance of electrons driven by an external power source. Our mission is to determine exactly how much current is required to produce $1\text{ mole}$ of these chromium ions in a specific timeframe of $48.25\text{ minutes}$.

The Half-Reaction

Before we can calculate anything, we must understand the stoichiometry of the reaction. The reduction of dichromate in an acidic medium is a classic redox half-reaction. Let's write it out:

Notice the crucial ratio here. To produce $2\text{ moles}$ of ${\text{Cr}}^{3+}$, the reaction demands exactly $6\text{ moles}$ of electrons. This is our golden key. By applying a simple unitary method, we can find the electron requirement for just $1\text{ mole}$ of ${\text{Cr}}^{3+}$:

So, to achieve our goal, we need to pump exactly $3\text{ moles}$ of electrons into the system.

Faraday's Law

Now, we transition from chemical moles to physical charge. How much electrical charge do $3\text{ moles}$ of electrons carry? This is where Michael Faraday's brilliant law comes into play. The total charge $Q$ is the product of the number of moles of electrons ($n_e$) and Faraday's constant ($F$), which is approximately $96500{\text{ C mol}}^{-1}$.

Substituting our values, we get:

We will leave this unmultiplied for now to make our final calculation smoother.

Bringing It All Together

In physics, electric charge is also defined as the product of current ($I$) and time ($t$).

A critical trap to avoid here is the unit of time. The standard SI unit for time in this equation is seconds, not minutes. We must convert the given time of $48.25\text{ minutes}$ into seconds:

Now, we equate our two expressions for the total charge $Q$:

Rearranging to solve for the current $I$:

Let's crunch the numbers. The numerator becomes $289500$, and the denominator simplifies beautifully to $2895$.

Dividing them yields a perfectly clean and satisfying answer: $100\text{ Amperes}$. This is a massive amount of current, highlighting the immense electrical energy required to drive industrial-scale electrochemical processes!