Problem 170
Question
The standard electrode potential of \(\mathrm{Cu}^{2} / \mathrm{Cu}=0.34 \mathrm{~V}\). The electrode potential will be zero, when the conc. of \(\mathrm{Cu}^{2+}\) is as \(\mathrm{x} \times 10^{-12} \mathrm{M}\). the value of \(\mathrm{x}\) is \([\) lig \(2=0.3010, \log 3=0.4771\) and \(\log 3.4=0.5315]\)
Step-by-Step Solution
Verified Answer
The value of \(x\) is 3.14.
1Step 1: Understand the Nernst Equation
The Nernst equation is used to calculate the electrode potential at any given concentration. It can be written as: \[ E = E^0 - \left( \frac{0.0591}{n} \right) \log [ \text{concentration of ions}] \]where \(E\) is the electrode potential, \(E^0\) is the standard electrode potential, \(n\) is the number of electrons transferred, and the concentration is in molarity (M).
2Step 2: Substitute Known Values into Nernst Equation
For the reduction reaction \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \), the standard electrode potential \(E^0\) is given as 0.34 V. We want \(E\) to be zero. Therefore:\[ 0 = 0.34 - \left( \frac{0.0591}{2} \right) \log [ \text{Cu}^{2+}] \]
3Step 3: Rearrange to Solve for Ion Concentration
Rearrange the equation to solve for \( \log [ \text{Cu}^{2+}] \):\[ \left( \frac{0.0591}{2} \right) \log [ \text{Cu}^{2+}] = 0.34 \]\[ \log [ \text{Cu}^{2+}] = \frac{0.34 \times 2}{0.0591} \]
4Step 4: Calculate Logarithm Value
Calculate the logarithmic value:\[ \log [ \text{Cu}^{2+}] = \frac{0.68}{0.0591} \approx 11.501 \]
5Step 5: Convert Logarithm to Concentration
Converting the logarithmic form to concentration form:\[ [ \text{Cu}^{2+}] = 10^{-11.501} \]This gives us the concentration as \(x \times 10^{-12}\).
6Step 6: Determine the Value of x
Since \( 10^{-11.501} \text{ can be written as } 3.14 \times 10^{-12} \), the value of \(x\) that corresponds to the concentration \([ \text{Cu}^{2+}] = x \times 10^{-12} \text{M} \) is 3.14.
Key Concepts
Nernst equationStandard electrode potentialConcentration of ions
Nernst equation
The Nernst equation is a fundamental concept in electrochemistry used to calculate the electrode potential of a cell when the ion concentration is not at standard conditions. This equation helps link the standard electrode potential to the actual conditions in a cell, such as ion concentration. It is expressed as:\[ E = E^0 - \left( \frac{0.0591}{n} \right) \log [ \text{ion concentration} ] \]**Key components of the Nernst equation include:**
- \(E\): The actual electrode potential at the given conditions.
- \(E^0\): The standard electrode potential measured under standard conditions (usually when all ions are at 1 M concentration).
- \(n\): The number of electrons transferred in the reaction.
- \([\text{ion concentration}]\): The concentration of the ions involved, which directly affects the electrode potential.
Standard electrode potential
The standard electrode potential, symbolized as \(E^0\), is the measure of the energy gained or lost when a chemical species undergoes reduction or oxidation under standard conditions. It is a critical component in determining the driving force of an electrochemical reaction. Standard conditions imply that all solutes are at 1 M concentration, gases at 1 atm pressure, and the temperature is 25°C.**Importance of Standard Electrode Potential:**
- \(E^0\) values help predict the direction of electron flow in redox reactions. Positive \(E^0\) values typically indicate a tendency to gain electrons and become reduced.
- Comparing \(E^0\) values for different half-reactions can help in determining which reactions are more likely to proceed spontaneously.
- They are used to calculate the overall cell potential for galvanic cells, which is pivotal in evaluating the capability of electrochemical cells to do work.
Concentration of ions
The concentration of ions within a solution is fundamental to the Nernst equation and impacts electrode potentials significantly. In electrochemistry, changing the concentration of ions can either enhance or diminish the driving force for electron movement, altering the electrochemical behavior of a system.
**Influence of Ion Concentration:**
- Increasing ion concentration can drive a reaction towards completion by shifting the equilibrium, often resulting in a higher electrode potential.
- Decreasing ion concentration has the opposite effect, reducing the electrode potential as fewer ions are available to participate in redox reactions.
- Logarithms in the Nernst equation emphasize that even small changes in concentration can significantly impact the cell potential.
Other exercises in this chapter
Problem 166
Match the following $$ \begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \begin{array}{l} \text { (a) } 50 \% \text { solution of } \m
View solution Problem 169
The current strength in ampere required to deposit \(8.0 \mathrm{~g}\) of silver in one hour is [At No. of \(\mathrm{Ag}=108\) ] approximately
View solution Problem 172
In the electrolysis of \(\mathrm{KI}, \mathrm{I}_{2}\) is formed at the anode by the reaction; \(2 \mathrm{I} \longrightarrow \mathrm{I}_{2}+2 \mathrm{e}^{-}\)
View solution Problem 175
Which of the following reaction is possible at anode? [2002] (a) \(\mathrm{F}_{2}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{~F}^{-}\) (b) \(2 \mathrm{H}^{+}+{
View solution