When during electrolysis of a solution of AgNO3, 9650 C of charge pass through the electroplating ba

Q: Why do we divide the charge by $96500$?

$96500\text{ C}$ is approximately the charge of one mole of electrons, known as Faraday's constant. Dividing the total charge by this constant gives us the number of moles of electrons that passed through the circuit.

Q: How do we know that one mole of electrons deposits one mole of silver?

By looking at the balanced half-reaction at the cathode: $\text{Ag}^+ + e^- \rightarrow \text{Ag}$. The stoichiometric coefficient of electrons is $1$, meaning one electron reduces one silver ion.

Q: What if the salt was $\text{CuSO}_4$ instead of $\text{AgNO}_3$?

For copper(II) sulfate, the reaction is $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$. Here, two moles of electrons are required to deposit one mole of copper. You would divide the moles of electrons by $2$ to get the moles of copper.

Q: Do we need to consider the anode reaction here?

No, the question specifically asks for the mass of silver deposited on the cathode. The anode reaction (which would be the oxidation of water or silver, depending on the electrode material) doesn't affect this calculation.

Q: Is the molar mass of silver always $108\text{ g/mol}$?

Yes, the standard atomic weight of silver is approximately $107.87\text{ u}$, which is commonly rounded to $108\text{ g/mol}$ for calculations in JEE and NEET unless specified otherwise.

Question

When during electrolysis of a solution of $	ext{AgNO}_3$, $9650	ext{ C}$ of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be

Options:

(A) $1.08	ext{ g}$

(B) $10.8	ext{ g}$

(C) $21.6	ext{ g}$

(D) $108	ext{ g}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of an electroplating bath. Inside the bath, there is a solution of silver nitrate ($	ext{AgNO}_3$). We are passing a total electrical charge of $9650	ext{ C}$ through this setup. Our main objective is to determine exactly how much silver gets deposited on the cathode.

This is a classic application of Faraday's First Law of Electrolysis, which relates the amount of substance deposited at an electrode to the quantity of electricity passed through the electrolyte.

### The Master Equation

Let's focus our attention on the cathode. Here, the positively charged silver ions ($	ext{Ag}^+$) from the solution migrate towards the negatively charged cathode. Each silver ion grabs one electron to become a solid silver atom. The chemical reaction can be written as:

$$ 	ext{Ag}^+ (aq) + e^- ightarrow 	ext{Ag} (s) $$

This stoichiometry tells us a very important fact: **one mole of electrons will deposit exactly one mole of silver**.

Now, how many moles of electrons did we actually pass through the circuit? To find this, we use Faraday's constant ($F \approx 96500	ext{ C/mol}$), which represents the charge of one mole of electrons. By dividing the total charge by Faraday's constant, we get the moles of electrons:

$$ n_{e^-} = \frac{Q}{F} = \frac{9650}{96500} $$

### Final Calculation

Calculating this fraction is quite simple. We find that exactly $0.1	ext{ moles}$ of electrons flowed through the circuit. Consequently, $0.1	ext{ moles}$ of silver must have been deposited on the cathode.

$$ n_{	ext{Ag}} = 0.1	ext{ mol} $$

Finally, to find the mass of the deposited silver, we multiply these moles by the molar mass of silver, which is $108	ext{ g/mol}$.

$$ m_{	ext{Ag}} = n_{	ext{Ag}} 	imes M_{	ext{Ag}} $$

$$ m_{	ext{Ag}} = 0.1 	imes 108 = 10.8	ext{ g} $$

This gives us our final answer: **$10.8	ext{ g}$** of silver is deposited on the cathode.

Always remember, if the question had mentioned a current efficiency (say, $80\%$), you would simply take $80\%$ of this theoretical mass to find the actual mass deposited. Keep an eye out for such twists!

Animated Solution for Chemistry - Electrochemistry: When during electrolysis of a solution of $AgNO_{3}$ , $9650 C$ of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be

Select Answer:

Visualized Solution

$E l ec t r o l y t i c C e l l S e t u p$

$C a t h o d e R e a c t i o n$

$M o l es o f E l ec t r o n s$

$C a l c u l a t in g M o l es$

$M a ss o f S i l v er D e p os i t e d$

$T h e W a y F or w a r d$

The Sigma Insight: Electrolytic Conduction

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

108 g of silver (molar mass $108 g mol^{- 1}$ ) is deposited at cathode from $AgNO_{3} (a q)$ solution by a certain quantity of electricity. The volume (in L) of oxygen gas produced at 273 K and 1 bar pressure from water by the same quantity of electricity is ………

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 = $63.5 u$ )

A solution of $Ni(NO_{3})_{2}$ is electrolysed between platinum electrodes using $0.1$ Faraday electricity. How many mole of Ni will be deposited at the cathode?

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

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electrolytic Cell Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Cathode Reaction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles of Electrons

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Moles

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass of Silver Deposited

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward