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If the 0.05 molar solution of M^+ is rep...

If the 0.05 molar solution of `M^+` is replaced by a 0.0025 Molar `M^+` solution. then the magnitude of the cell potential would be :

A

35mV

B

70mV

C

140mV

D

700mV

Text Solution

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The correct Answer is:
To find the magnitude of the cell potential when the concentration of \( M^+ \) is changed from 0.05 M to 0.0025 M, we can use the Nernst equation for a concentration cell. Here’s the step-by-step solution: ### Step 1: Understand the Nernst Equation The Nernst equation is given by: \[ E_{cell} = E^0 - \frac{RT}{nF} \ln Q \] Where: - \( E_{cell} \) is the cell potential, - \( E^0 \) is the standard cell potential, - \( R \) is the universal gas constant (8.314 J/(mol·K)), - \( T \) is the temperature in Kelvin, - \( n \) is the number of moles of electrons transferred, - \( F \) is Faraday's constant (96485 C/mol), - \( Q \) is the reaction quotient. ### Step 2: Identify the Initial and Final Concentrations We have: - Initial concentration of \( M^+ \) (C1) = 0.05 M - Final concentration of \( M^+ \) (C2) = 0.0025 M ### Step 3: Calculate the Reaction Quotient (Q) For a concentration cell, the reaction quotient \( Q \) can be expressed as: \[ Q = \frac{[M^+]_{anode}}{[M^+]_{cathode}} \] In this case, we can assume: - At the anode: \( [M^+] = 0.0025 \, M \) - At the cathode: \( [M^+] = 0.05 \, M \) Thus, \[ Q = \frac{0.0025}{0.05} = 0.05 \] ### Step 4: Calculate the Cell Potential Change Using the Nernst equation, we can express the change in cell potential as: \[ E_{cell} = E^0 - \frac{0.059}{n} \log Q \] Since we are comparing two states, we can set up the ratio: \[ \frac{E_{cell2}}{E_{cell1}} = \frac{\log [M^+]_{2}}{\log [M^+]_{1}} \] Substituting the values: \[ \frac{E_{cell2}}{E_{cell1}} = \frac{\log(0.0025)}{\log(0.05)} \] ### Step 5: Calculate the Logarithmic Values Calculating the logarithmic values: \[ \log(0.0025) \approx -2.6 \] \[ \log(0.05) \approx -1.3 \] ### Step 6: Calculate the Ratio of Cell Potentials Now, substituting these values into the ratio: \[ \frac{E_{cell2}}{70 \, \text{mV}} = \frac{-2.6}{-1.3} = 2 \] Thus, \[ E_{cell2} = 2 \times 70 \, \text{mV} = 140 \, \text{mV} \] ### Final Answer The magnitude of the cell potential when the concentration of \( M^+ \) is changed to 0.0025 M is **140 mV**. ---

To find the magnitude of the cell potential when the concentration of \( M^+ \) is changed from 0.05 M to 0.0025 M, we can use the Nernst equation for a concentration cell. Here’s the step-by-step solution: ### Step 1: Understand the Nernst Equation The Nernst equation is given by: \[ E_{cell} = E^0 - \frac{RT}{nF} \ln Q \] Where: ...
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