Q52 / 180 · KCET 2025 · Chemistry

Match the following select the correct option for the quantity of electricity, in $\mathrm{Cmol}^{-1}$ required to deposit various metals at cathode \begin{array}{|l|l|l|l|}\hline & List - I & & List- II \\\hline a & \mathrm{Ag}^{+} & i & 386000 \mathrm{Cmol}^{-1} \\\hline b & \mathrm{Mg}^{2+} & ii & 289500 \mathrm{Cmol}^{-1} \\\hline c & \mathrm{Al}^{3+} & iii & 96500 \mathrm{Cmol}^{-1} \\\hline d & \mathrm{Ti}^{4+} & iv & 193000 \mathrm{Cmol}^{-1} \\\hline\end{array}

  • A
    $a-i i, b-i, c-i v, d-i i i$
  • Correct answer
    B
    $a-i i i, b-i v, c-i i, d-i$
  • C
    $a-i v, b-i i i, c-i, d-i i$
  • D
    $a-i, b-i i, c-i i i, d-i v$

Solution

$\begin{aligned} 1 \mathrm{~F}=96500 \mathrm{C} & \\ \mathrm{Ag}^{+}+\mathrm{e}^{-} & \rightarrow \mathrm{Ag}\left[\frac{\mathrm{C}}{\mathrm{~mol}}=1 \times 96500 \mathrm{C}=96500 \mathrm{C}\right] \\ 1 \mathrm{~F} & \\ \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} & \rightarrow \mathrm{Mg}\left[\frac{\mathrm{C}}{\mathrm{~mol}}=2 \times 96500 \mathrm{C}=193000\right. \\ 2 \mathrm{~F} & \\ \mathrm{Al}^{3+}+3 \mathrm{e}^{-} & \rightarrow \mathrm{Al}\left[\frac{\mathrm{C}}{\mathrm{~mol}}=3 \times 96500 \mathrm{C}=289500\right] \\ 3 \mathrm{~F} & \\ \mathrm{Ti}^{4+}+4 \mathrm{e}^{-} & \rightarrow \mathrm{Ti}\left[\frac{\mathrm{C}}{\mathrm{~mol}}=4 \times 96500 \mathrm{C}=386000\right] \\ 4 \mathrm{~F} & \end{aligned}$

Hence, the correct answer is option: 'B'.

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