Import Question JSON

Current Question (ID: 18174)

Question:

$\text{Cu(s)}|\text{Cu}^{+2}(10^{-3} \text{ M}) || \text{Ag}^+(10^{-5} \text{ M})|\text{Ag(s)}$ $\text{if } E^\circ_{\text{Cu}^{+2}/\text{Cu}} = +0.34 \text{ V, and } E^\circ_{\text{Ag}^+/\text{Ag}} = +0.80 \text{ V}$ $E_{\text{cell}} \text{ will be:}$

Options:

1. $0.46 \text{ V}$
2. $0.46 - \frac{RT}{2F} \ln 10^7$
3. $0.46 + \frac{RT}{2F} \ln 10^7$
4. $0.46 - \frac{RT}{2F} \ln 10^2$

Solution:

$\text{Step 1:}$ $\text{The anode and cathode reactions are as follows:}$ $\text{Anode}$ $\text{Cu} \rightarrow \text{Cu}^{+2} + 2e^- \quad \text{....(1)}$ $\text{Cathode}$ $2 \text{Ag} + 2e^- \rightarrow 2 \text{Ag} \quad \text{....(2)}$ $\text{Cell reaction is as follows:}$ $\text{Cu} + 2 \text{Ag}^+ \rightarrow \text{Cu}^{+2} + 2 \text{Ag(s)}$ $Q = \frac{[\text{Cu}^{+2}]}{[\text{Ag}^+]^2}$ $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q$ $\text{Step 2:}$ $\text{Calculate the value of } E_{\text{cell}} \text{ as follows:}$ $E_{\text{cell}} = ( -0.39 + 0.80 ) - \frac{RT}{nF} \ln \frac{[\text{Cu}^{+2}]}{[\text{Ag}^+]^2}$ $= +0.46 - \frac{RT}{2F} \ln \frac{[\text{Cu}^{+2}]}{[\text{Ag}^+]^2}$ $= +0.46 - \frac{RT}{2F} \ln \frac{10^{-3}}{(10^{-5})^2}$ $= +0.46 - \frac{RT}{2F} \ln 10^{-3} \times 10^{+10}$ $= 0.46 - \frac{RT}{2F} \ln 10^7$

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
}