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Current Question (ID: 18836)

Question:

$\text{For M}^{2+}/\text{M, and M}^{3+}/\text{M}^{2+}, \text{ the } E^\circ \text{ values of some metals are given in the table below:}$ $\begin{array}{|c|c|} \hline \text{Cr}^{+2}/\text{Cr} = -0.9 \text{ V} & \text{Cr}^{+3}/\text{Cr}^{+2} = -0.4 \text{V} \\ \hline \text{Mn}^{+2}/\text{Mn} = -1.2 \text{V} & \text{Mn}^{+3}/\text{Mn}^{+2} = +1.5 \text{V} \\ \hline \text{Fe}^{+2}/\text{Fe} = -0.4 \text{V} & \text{Fe}^{+3}/\text{Fe}^{+2} = +0.8 \text{V} \\ \hline \end{array}$ $\text{The correct inference(s) from the above data is/are:}$

Options:

1. $\text{The increasing order of their stability is Mn}^{3+}< \text{Fe}^{3+}< \text{Cr}^{3+}$
2. $\text{The increasing order of the above-mentioned metals ability to get oxidized is: Fe < Cr < Mn}$
3. $\text{The increasing order of the above-mentioned metals ability to get oxidized is: Fe > Cr > Mn}$
4. $\text{Both 1 and 2}$

Solution:

$\text{STEP 1: The } E^\circ \text{ value for Fe}^{+3}/\text{Fe}^{+2} \text{ is higher than that for Cr}^{+3}/\text{Cr}^{+2} \text{ and lower than that for Mn}^{+3}/\text{Mn}^{+2}. $ $\text{So, the reduction of Fe}^{+3} \text{ to Fe}^{+2} \text{ is easier than the reduction of Mn}^{+3} \text{ to Mn}^{2+}, \text{ but not as easy as the reduction of Cr}^{+3} \text{ to Cr}^{2+}. $ $\text{Hence, Fe}^{+3} \text{ is more stable than Mn}^{+3}, \text{ but less stable than Cr.}$ $\text{These metal ions can be arranged in the increasing order of their stability as: Mn}^{+3} < \text{Fe}^{+3} < \text{Cr}^{+3}$ $\text{STEP 2: The reduction potentials for the given pairs increase in the following order. Mn}^{+2}/\text{Mn} < \text{Cr}^{+2}/\text{Cr} < \text{Fe}^{+2}/\text{Fe}.$ $\text{So, the oxidation of Fe to Fe}^{+2} \text{ is not as easy as the oxidation of Cr to Cr}^{2+} \text{ and the oxidation of Mn to Mn}^{2+}. $ $\text{Thus, these metals can be arranged in the increasing order of their ability to get oxidised as: Fe < Cr < Mn}$

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{
  "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
}