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Current Question (ID: 18155)

Question:
$\text{Consider the following relations for emf of an electrochemical cell:}$ $\text{(a) emf of cell} = (\text{Oxidation potential of anode}) - (\text{Reduction potential of cathode})$ $\text{(b) emf of cell} = (\text{Oxidation potential of anode}) + (\text{Reduction potential of cathode})$ $\text{(c) emf of cell} = (\text{Reduction potential of anode}) + (\text{Reduction potential of cathode})$ $\text{(d) emf of cell} = (\text{Oxidation potential of anode}) - (\text{Oxidation potential of cathode})$ $\text{The correct relation among the given options is:}$
Options:
  • 1. $\text{(a) and (b)}$
  • 2. $\text{(c) and (d)}$
  • 3. $\text{(b) and (d)}$
  • 4. $\text{(c) and (a)}$
Solution:
$\text{HINT: Emf of cell} = E_{\text{ox}} + E_{\text{red}}$ $\text{Explanation:}$ $\text{Emf of a cell} = (\text{Oxidation potential of anode}) + (\text{Reduction potential of cathode})$ $= (\text{Oxidation potential of anode}) + (-\text{oxidation potential of cathode})$ $= (\text{Oxidation potential of anode}) - (\text{Oxidation potential of cathode})$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}