Import Question JSON

Current Question (ID: 18176)

Question:

$\text{Fe}^{2+}(\text{aq}) + \text{Ag}^+(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{Ag}(\text{s})$ $E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.77 \text{ V}; \; E^\circ_{\text{Ag}^+/\text{Ag}} = 0.80 \text{ V}$ $\text{The value of } \Delta G_r^\circ \text{ in the above reaction will be:}$

Options:

1. $+2.89 \text{ kJ}$
2. $-2.89 \text{ kJ}$
3. $-2.89 \text{ J}$
4. $-186.83 \text{ J}$

Solution:

$\text{Hint: } \Delta_t G^\circ = -nF E_\text{cell}^\circ$ $\text{Step 1:}$ $\text{Calculate the value of } E_\text{cell}^\circ \text{ is as follows:}$ $E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.77 \text{ V}$ $E^\circ_{\text{Ag}^+/\text{Ag}} = 0.80 \text{ V}$ $\text{The galvanic cell of the given reaction is depicted as:}$ $\text{Fe}^{2+}(\text{aq})|\text{Fe}^{3+}(\text{aq})||\text{Ag}^+(\text{aq})|\text{Ag}(\text{s})$ $\text{Now, the standard cell potential is}$ $E_\text{cell}^\circ = E_\text{R}^\circ - E_\text{L}^\circ$ $= 0.80 - 0.77$ $= 0.03 \text{ V}$ $\text{Here, } n=1$ $\text{Step 2:}$ $\text{Then, } \Delta_t G^\circ = -nF E_\text{cell}^\circ$ $= -1 \times 96487 \text{ C mol}^{-1} \times 0.03 \text{ V}$ $= -2894.61 \text{ J mol}^{-1}$ $= -2.89 \text{ kJ mol}^{-1}$

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
}