Import Question JSON

Current Question (ID: 18845)

Question:

$\text{Zinc can be coated on iron to produce galvanised iron, but the reverse is not possible because:}$

Options:

1. $\text{Zinc is lighter than iron.}$
2. $\text{Zinc has a lower melting point than iron.}$
3. $\text{Zinc has a lower negative reduction potential than iron.}$
4. $\text{Zinc has a higher negative reduction potential than iron.}$

Solution:

$\text{HINT: Zinc has a higher negative reduction potential than iron.}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{The metal with more negative standard reduction potential, have higher tendency to get oxidised.}$ $\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \; ; \; E^0 = -0.76 \text{ V}$ $\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \; ; \; E^0 = -0.44 \text{ V}$ $\text{In galvanized iron, Zn has higher negative reduction potential (i.e. less RP or more OP) Zn lose electrons because of more}$ $\text{tendency to oxidise.}$ $\text{STEP 2:}$ $\text{Thus, Zn works as anode and protects iron from rusting by making iron as cathode.}$ $\text{The reverse is not possible because even if}$ $\text{the zinc is coated with iron means iron will be in contact with moisture, zinc will get}$ $\text{oxidized first due to high negative reduction potential.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}