Import Question JSON

Current Question (ID: 8629)

Question:

$\text{E}^\Theta \text{ values of some redox couples are given below. On the basis of these values choose the correct option.}$ $\text{E}^\Theta \text{ values : Br}_2/\text{Br}^- = +1.90$ $\text{Ag}^+/\text{Ag}_{(\text{s})} = +0.80$ $\text{Cu}^{2+}/\text{Cu}_{(\text{s})} = +0.34;$ {I}_2_{(\text{s})}/\text{I}^- = +0.54$

Options:

1. $\text{Cu will reduce Br}^-$
2. $\text{Cu will reduce Ag}$
3. $\text{Cu will reduce I}^-$
4. $\text{Cu will reduce Br}_2$

Solution:

$\text{Hint: High reduction potential substance is a good oxidising agent}$ $\text{The given values of E}^\circ \text{ are as follows:}$ $\text{Br}_2/\text{Br}^- = +1.90\text{V}$ $\text{Ag}/\text{Ag}^+ = -0.80\text{V}$ $\text{Cu}^{2+}/\text{Cu}_{(\text{s})} = +0.34 {V}$ $\text{I}^-/\text{I}_2_{(\text{s})} = -0.54\text{ V}$ $\text{Br}^-/\text{Br}_2 = -1.90\text{V}$ $\text{The E}^\circ \text{ values show that copper will reduce Br}_2, \text{ if the E}^\circ \text{ of the following redox reaction is positive.}$ $\underline{2\text{Cu} + \text{Br}_2 \rightarrow \text{CuBr}_2}$ $\text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{e}^- ; \text{E}^\circ = -0.34\text{V}$ $\text{Br}_2 + 2\text{e}^- \rightarrow 2\text{Br}^- ; \text{E} = +1.09\text{V Cu} + \text{Br}_2 \rightarrow \text{CuBr}_2 ; \text{E}^\circ = +0.75$ $\text{Since, E}^\circ \text{ of this reaction is positive, therefore, Cu can reduce Br}_2.$ $\text{While other reaction will give negative value.}$

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
}