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Current Question (ID: 18223)

Question:

$\text{For the cell, Ti/Ti}^{+}(0.001\text{M})||\text{Cu}^{2+}(0.1\text{M})|\text{Cu}, E^\circ_{\text{cell}}$ $\text{at } 25\ ^\circ \text{C is } 0.83\ \text{V. } E_{\text{cell}} \text{ can be increased:}$

Options:

1. $\text{By increasing } [\text{Cu}^{2+}]$
2. $\text{By increasing } [\text{Ti}^{+}]$
3. $\text{By decreasing } [\text{Cu}^{2+}]$
4. $\text{None of the above.}$

Solution:

$\text{HINT: } E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \frac{[\text{Ti}^{+}]^2}{[\text{Cu}^{2+}]}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{Ti / Ti}^{+} (0.001\text{M})/1\text{Cu}^{2+} (0.1\ \text{M})/\text{Cu}$ $\text{Ti} \rightarrow \text{Ti}^{+} + \text{e}^{-} \times 2$ $\text{Cu}^{2+} + 2\text{e}^{-} \rightarrow \text{Cu}$ $2\ \text{Ti} + \text{Cu}^{2+} \rightarrow 2\ \text{Ti}^{+} + \text{Cu}$ $\text{For this cell Nernst equation is as follows:}$ $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \frac{[\text{Ti}^{+}]^2}{[\text{Cu}^{2+}]}$ $\text{STEP 2:}$ $\text{If } \text{Cu}^{2+} \text{ ion concentration } \uparrow\uparrow\uparrow \text{ then:}$ $\frac{0.059}{n} \log \frac{[\text{Ti}^{+}]^2}{[\text{Cu}^{2+}]} \downarrow\downarrow\downarrow$ $\text{but overall } E^\circ_{\text{cell}} - \frac{0.059}{n} \uparrow \text{ increases. Hence, } E_{\text{cell}} \text{ increases}$ $\text{So on increasing } \text{Cu}^{2+} \text{ ion concentration.}$ $E_{\text{cell}} \text{ increases.}$

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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
}