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Current Question (ID: 8623)

Question:
$\text{The Mn}^{3+} \text{ ion is unstable in solution and undergoes disproportionation reaction to give Mn}^{2+}\text{, MnO}_2 \text{ and H}^+ \text{ ion. The balanced ionic equation for the reaction is-}$
Options:
  • 1. $2\text{Mn}^{3+}_{(aq)} + 2\text{H}_2\text{O}_{(l)} \rightarrow \text{MnO}_{2(s)} + \text{Mn}^{2+}_{(aq)} + 4\text{H}^+_{(aq)}$
  • 2. $\text{Mn}^{3+}_{(aq)} + \text{H}_2\text{O}_{(l)} \rightarrow \text{MnO}_{2(s)} + 2\text{Mn}^{2+}_{(aq)} + 4\text{H}^+_{(aq)}$
  • 3. $5\text{Mn}^{3+}_{(aq)} + 2\text{H}_2\text{O}_{(l)} \rightarrow \text{MnO}_{2(s)} + 3\text{Mn}^{2+}_{(aq)} + 4\text{H}^+_{(aq)}$
  • 4. $2\text{Mn}^{3+}_{(aq)} + 2\text{H}_2\text{O}_{(l)} \rightarrow 2\text{MnO}_{2(s)} + 2\text{Mn}^{2+}_{(aq)} + 4\text{H}^+_{(aq)}$
Solution:
$\text{Hint: Left-hand side and right-hand side of the reaction overall charged must be balanced}$ $\text{The given reaction can be represented as:}$ $\text{Mn}^{3+}_{(aq)} \rightarrow \text{Mn}^{2+}_{(aq)} + \text{MnO}_{2(g)} + \text{H}^+_{(aq)}$ $\text{The oxidation half equation is:}$ $\stackrel{+3}{\text{Mn}}^{3+}_{(aq)} \rightarrow \stackrel{+4}{\text{MnO}}_2_{(s)}$ $\text{The oxidation number is balanced by adding one electron as:}$ $\text{Mn}^{3+}_{(aq)} \rightarrow \text{MnO}_{2(s)} + e^-$ $\text{The charge is balanced by adding 4H}^+ \text{ ions as:}$ $\text{Mn}^{3+}_{(aq)} \rightarrow \text{MnO}_{2(s)} + 4\text{H}^+_{(aq)} + e^-$ $\text{The O atoms and H}^+ \text{ ions are balanced by adding 2H}_2\text{O molecules as:}$ $\text{Mn}^{3+}_{(aq)} + 2\text{H}_2\text{O}_{(l)} \rightarrow \text{MnO}_{2(s)} + 4\text{H}^+_{(aq)} + e^- \text{ .......(i)}$ $\text{The reduction half equation is:}$ $\text{Mn}^{3+}_{(aq)} \rightarrow \text{Mn}^{2+}_{(aq)}$ $\text{The oxidation number is balanced by adding one electron as:}$ $\text{Mn}^{3+}_{(aq)} + e^- \rightarrow \text{Mn}^{2+}_{(aq)} \text{ .......(ii)}$ $\text{The balanced chemical equation can be obtained by adding equations (i) and (ii) as:}$ $2\text{Mn}^{3+}_{(aq)} + 2\text{H}_2\text{O}_{(l)} \rightarrow \text{MnO}_{2(s)} + \text{Mn}^{2+}_{(aq)} + 4\text{H}^+_{(aq)}$

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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
}