What are fuel cells? State Two advantages of hydrogen-oxygen fuel cell over ordinary cell.
What are fuel cells? State Two advantages of hydrogen-oxygen fuel cell over ordinary cell.
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Solution and Explanation
Step 1: Definition of Fuel Cells: Fuel cells are electrochemical devices that convert chemical energy into electrical energy through a redox reaction between fuel (e.g., hydrogen) and an oxidizing agent (e.g., oxygen).
Step 2: Advantages of Hydrogen-Oxygen Fuel Cell: 1. Higher Efficiency: Fuel cells have higher efficiency compared to conventional electrochemical cells. 2. Eco-friendly: Produces only water as a byproduct, making it environmentally friendly.
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
- Chemistry
- Electrochemistry
- (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}
- CBSE CLASS XII - 2025
- Chemistry
- Electrochemistry
(a.)Write the anode and cathode reactions and the overall cell reaction occurring in a lead storage battery during its use.
- CBSE CLASS XII - 2025
- Chemistry
- Electrochemistry
- The charge required for the reduction of 1 mol of \( MnO_4^- \) to \( MnO_2 \) is
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- Chemistry
- Electrochemistry
- 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
- Chemistry
- Electrochemistry
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