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Current Question (ID: 18253)

Question:

$\text{The negative sign in the expression } E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V indicates:}$ $1. \text{ The reactivity of the metal increases.}$ $2. \text{ The reactivity of the metal decreases.}$ $3. \text{ There is no effect on the metal's reactivity.}$ $4. \text{ None of the above.}$

Options:

1. $\text{The reactivity of the metal increases.}$
2. $\text{The reactivity of the metal decreases.}$
3. $\text{There is no effect on the metal's reactivity.}$
4. $\text{None of the above.}$

Solution:

$\text{HINT: More -ve reduction potential suggest more reactivity than hydrogen and it can act as anode.}$ $\text{Explanation:}$ $\text{Greater the negative sign of standard reduction potential of metal greater is the reactivity.}$ $\text{It means that Zn is more reactive than hydrogen.}$ $\text{When zinc electrode will be connected to standard hydrogen electrode, Zn will get oxidised and H}^+ \text{ will get reduced.}$ $\text{Thus, zinc electrode will be the anode of the cell and hydrogen electrode will be the cathode of the cell.}$

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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
}