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Current Question (ID: 8622)

Question:

\text{Standard reduction potentials of the half-reactions are given below:} \text{F}_2\text{(g)} + 2e^- \rightarrow 2\text{F}^-\text{(aq)} ; E^\circ = +2.85 \text{ V} \text{Cl}_2\text{(g)} + 2e^- \rightarrow 2\text{Cl}^-\text{(aq)} ; E^\circ = +1.36 \text{ V} \text{Br}_2\text{(g)} + 2e^- \rightarrow 2\text{Br}^-\text{(aq)} ; E^\circ = +1.06 \text{ V} \text{I}_2\text{(g)} + 2e^- \rightarrow 2\text{I}^-\text{(aq)} ; E^\circ = +0.53 \text{ V} \text{The strongest oxidizing and reducing agents, respectively, are:}

Options:

1. $\text{Br}_2 \text{ and Cl}^-$
2. $\text{Cl}_2 \text{ and Br}^-$
3. $\text{Cl}_2 \text{ and I}_2$
4. $\text{F}_2 \text{ and I}^-$

Solution:

$\text{HINT : More reduction potential, better oxidising agent. Less reduction potential, better reducing agent.}$ $\text{The oxidizing agent is a substance that causes oxidation by accepting electrons; therefore, its oxidation state decreases. The reducing agent is a substance that causes reduction by losing electrons; therefore its oxidation state increases.}$ $\text{On the basis of reduction potential, the strength of oxidising and reducing nature can be understood.}$ $\text{More positive reduction potential means the substance will get reduce easily and will oxidise other. So, it will be better oxidising agent and vice versa for reducing agent.}$ $\text{In the given data all are Standard reduction potentials (SRP). So, best oxidising agent is F}_2 \text{ and best reducing agent is I}^-\text{.}$

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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
}