Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at th

Q: What exactly is a Faraday?

A Faraday ($F$) is a unit of electrical charge. It represents the total charge carried by exactly one mole of electrons, which is approximately $96,500$ Coulombs.

Q: Why do we need 2 Faradays to deposit 1 mole of copper?

Copper in copper sulphate exists as $\text{Cu}^{2+}$ ions. To reduce one $\text{Cu}^{2+}$ ion to a neutral $\text{Cu}$ atom, it must gain 2 electrons. Therefore, 1 mole of $\text{Cu}^{2+}$ requires 2 moles of electrons, which equals 2 Faradays of charge.

Q: What would happen if we passed only 1 Faraday of electricity?

Since 2 Faradays deposit 1 mole of copper, passing only 1 Faraday would deposit exactly half a mole of copper, which is $31.75 \text{ g}$.

Q: Does the concentration of the $\text{CuSO}_4$ solution affect the mass deposited?

As long as there are enough $\text{Cu}^{2+}$ ions available to be reduced, the mass deposited depends *only* on the total charge passed, according to Faraday's First Law of Electrolysis.

Q: How do we know the reaction happens at the cathode and not the anode?

Reduction (gain of electrons) always occurs at the cathode. Since $\text{Cu}^{2+}$ ions are positively charged, they migrate to the negatively charged cathode to gain electrons and deposit as solid metal.

Question

Two Faraday of electricity is passed through a solution of $	ext{CuSO}_4$. The mass of copper deposited at the cathode is (at. mass of Cu = $63.5	ext{ u}$)

Options:

(A) $0	ext{ g}$

(B) $63.5	ext{ g}$

(C) $2	ext{ g}$

(D) $127	ext{ g}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine an electrolytic cell filled with a vibrant blue solution of copper sulphate ($	ext{CuSO}_4$). When we pass an electric current through this solution, a fascinating chemical dance begins. The copper sulphate dissociates into copper ions ($	ext{Cu}^{2+}$) and sulphate ions ($	ext{SO}_4^{2-}$).

The positively charged copper ions are naturally drawn towards the negatively charged electrode, which is the cathode. This is where the magic of electroplating happens!

### The Master Equation
To figure out exactly how much copper gets deposited, we need to look closely at the reduction reaction taking place at the cathode.

$$ 	ext{Cu}^{2+} + 2e^- ightarrow 	ext{Cu}_{(s)} $$

This simple equation is our master key. It tells us that every single copper ion requires exactly two electrons to neutralize its charge and become a solid atom of copper. Scaling this up to the macroscopic world, it means that to deposit **one mole** of copper metal, we must supply exactly **two moles** of electrons.

### The Power of Faraday
Now, let's bring in Michael Faraday's brilliant insight. The total electrical charge carried by one mole of electrons is defined as one Faraday ($1F$).

Since our balanced equation demands two moles of electrons for every mole of copper, it translates directly to:
**$2F$ of electricity deposits $1$ mole of Copper.**

### Final Calculation
The problem states that we are passing exactly $2F$ of electricity through the solution. Based on our stoichiometry, this is the exact amount needed to deposit precisely one mole of copper at the cathode.

All that's left is to convert this molar amount into a physical mass. The atomic mass of copper is given as $63.5 	ext{ u}$, which means one mole of copper weighs $63.5 	ext{ g}$.

Therefore, the mass of copper deposited is **$63.5 	ext{ g}$**.

Animated Solution for Chemistry - Electrochemistry: Two Faraday of electricity is passed through a solution of $CuSO_{4}$ . The mass of copper deposited at the cathode is (at. mass of Cu = $63.5 u$ )

Select Answer:

Visualized Solution

\text{Electrolytic Cell Setup}

\text{Dissociation of } \text{CuSO}_4

\text{Cathode Reaction}

\text{Stoichiometry}

\text{Faraday\'s Law}

\text{Final Calculation}

\text{Conclusion}

The Sigma Insight: Electrolytic Conduction

Analyzing the Setup Imagine an electrolytic cell filled with a vibrant blue solution of copper sulphate ( $CuSO_{4}$ )

The Master Equation

The Power of Faraday Now, let's bring in Michael Faraday's brilliant insight

Final Calculation The problem states that we are passing exactly $2 F$ of electricity through the solution

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