How many Seconds are reQuircd to deposst o.2859 Or Cadmium metal from a Solutior that Contams cazt ions if a Current or Oileos A is applied n And 225 g or Magnesium metal in 30.n Seconds fram a Solutiorn That (ontams Mg?t ions A तपार एरे एुने
How many Seconds are reQuircd to deposst o.2859 Or Cadmium metal from a Solutior that Contams cazt ions if a Current or Oileos A is applied n And 225 g or Magnesium metal in 30.n Seconds fram a Solutiorn That (ontams Mg?t ions A तपार एरे एुने
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Electrochemistry Calculation Problems
**Problem 1:**
How many seconds are required to deposit 0.285 g of Cadmium metal from a solution that contains \( Cd^{2+} \) ions if a current of 0.105 A is applied?
\[ \boxed{\text{___} \text{S}} \]
**Problem 2:**
And, 0.225 g of Magnesium metal in 301 seconds from a solution that contains \( Mg^{2+} \) ions?
\[ \boxed{\text{A}} \]
### Explanation
These problems require the use of Faraday's laws of electrolysis to solve for the time or current needed to deposit a certain amount of metal from a solution.
#### Problem 1: Steps for Solution
1. **Calculate the moles of Cadmium deposited:**
- Molar Mass of \(Cd\): 112.4 g/mol
- Moles of Cd \((n)\): \(\frac{0.285 \text{ g}}{112.4 \text{ g/mol}}\)
2. **Calculate the total charge (Q) used:**
- \(Cd^{2+}\) ions require 2 moles of electrons per mole of \(Cd\) deposited.
- Faraday's constant (\(F\)): 96485 C/mol
- Total charge (C): \(n \times 2 \times F\)
3. **Calculate time required:**
- Current (I): 0.105 A
- Time (t): \(\frac{Q}{I}\)
#### Problem 2: Steps for Solution
1. **Calculate the moles of Magnesium deposited:**
- Molar Mass of \(Mg\): 24.3 g/mol
- Moles of Mg \((n)\): \(\frac{0.225 \text{ g}}{24.3 \text{ g/mol}}\)
2. **Calculate the total charge (Q) used:**
- \(Mg^{2+}\) ions require 2 moles of electrons per mole of \(Mg\) deposited.
- Faraday's constant (\(F\)): 96485 C/mol
- Total charge (C): \(n \times 2 \times F\)
3. **Calculate current (I):**
- Time (t): 301 s
- Current](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcfcc845b-36c3-4587-86f5-d581ad6d0ef3%2F2e4ed1f1-2b8c-4830-b44f-6d7387069305%2Fc4yyvx.jpeg&w=3840&q=75)
Transcribed Image Text:### Electrochemistry Calculation Problems
**Problem 1:**
How many seconds are required to deposit 0.285 g of Cadmium metal from a solution that contains \( Cd^{2+} \) ions if a current of 0.105 A is applied?
\[ \boxed{\text{___} \text{S}} \]
**Problem 2:**
And, 0.225 g of Magnesium metal in 301 seconds from a solution that contains \( Mg^{2+} \) ions?
\[ \boxed{\text{A}} \]
### Explanation
These problems require the use of Faraday's laws of electrolysis to solve for the time or current needed to deposit a certain amount of metal from a solution.
#### Problem 1: Steps for Solution
1. **Calculate the moles of Cadmium deposited:**
- Molar Mass of \(Cd\): 112.4 g/mol
- Moles of Cd \((n)\): \(\frac{0.285 \text{ g}}{112.4 \text{ g/mol}}\)
2. **Calculate the total charge (Q) used:**
- \(Cd^{2+}\) ions require 2 moles of electrons per mole of \(Cd\) deposited.
- Faraday's constant (\(F\)): 96485 C/mol
- Total charge (C): \(n \times 2 \times F\)
3. **Calculate time required:**
- Current (I): 0.105 A
- Time (t): \(\frac{Q}{I}\)
#### Problem 2: Steps for Solution
1. **Calculate the moles of Magnesium deposited:**
- Molar Mass of \(Mg\): 24.3 g/mol
- Moles of Mg \((n)\): \(\frac{0.225 \text{ g}}{24.3 \text{ g/mol}}\)
2. **Calculate the total charge (Q) used:**
- \(Mg^{2+}\) ions require 2 moles of electrons per mole of \(Mg\) deposited.
- Faraday's constant (\(F\)): 96485 C/mol
- Total charge (C): \(n \times 2 \times F\)
3. **Calculate current (I):**
- Time (t): 301 s
- Current
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