Explain the difference between an electrochemical cell and electrolytic cell. State Kohlrausch's law and write its applications.
Explain the difference between an electrochemical cell and electrolytic cell. State Kohlrausch's law and write its applications.
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Solution and Explanation
Step 1: Difference Between Electrochemical and Electrolytic Cells: Feature & Electrochemical Cell & Electrolytic Cell
Energy Conversion & Chemical to electrical & Electrical to chemical
Spontaneity & Spontaneous reaction & Non-spontaneous reaction
Example & Galvanic/Daniel cell & Electrolysis of NaCl
Step 2: Kohlrausch's Law: \[ \Lambda_m^0 = \lambda^0_+ + \lambda^0_- \] where \( \lambda^0_+ \) and \( \lambda^0_- \) are the limiting molar conductivities of cations and anions, respectively.
Step 3: Applications: - Used to calculate the molar conductivity of weak electrolytes. - Helps in determining dissociation constants of weak acids and bases.
Top Questions on Electrochemistry
Standard electrode potential for \( \text{Sn}^{4+}/\text{Sn}^{2+} \) couple is +0.15 V and that for the \( \text{Cr}^{3+}/\text{Cr} \) couple is -0.74 V. The two couples in their standard states are connected to make a cell. The cell potential will be:
To calculate the cell potential (\( E^\circ_{\text{cell}} \)), we use the standard electrode potentials of the given redox couples.
Given data:
\( E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +0.15V \)
\( E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74V \)- CBSE CLASS XII - 2025
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- (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K: \[ \text{Sn}(s) \mid \text{Sn}^{2+} (\text{0.004 M}) \parallel \text{H}^+ (\text{0.02 M}) \mid \text{H}_2 (\text{1 Bar}) \mid \text{Pt}(s) \] (Given: \( E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.14 \, \text{V}, E^\circ_{\text{H}^+/\text{H}_2} = 0.00 \, \text{V} \)) \vspace{5pt}
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(a.)Write the anode and cathode reactions and the overall cell reaction occurring in a lead storage battery during its use.
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- The charge required for the reduction of 1 mol of \( MnO_4^- \) to \( MnO_2 \) is
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- Calculate \( \Lambda_m^0 \) for acetic acid and its degree of dissociation (\( \alpha \)) if its molar conductivity is 48.1 \( \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).
Given that
\( \Lambda_m^0 (\text{HC}) = 426 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \),
\( \Lambda_m^0 (\text{NaCl}) = 126 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \),
\( \Lambda_m^0 (\text{CH}_3\text{COONa}) = 91 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).- CBSE CLASS XII - 2025
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