JEE MAIN - Chemistry (2023 - 30th January Morning Shift - No. 18)
Consider the cell
$$\mathrm{Pt}_{(\mathrm{s})}\left|\mathrm{H}_{2}(\mathrm{~g}, 1 \mathrm{~atm})\right| \mathrm{H}^{+}(\mathrm{aq}, 1 \mathrm{M})|| \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq}) \mid \operatorname{Pt}(\mathrm{s})$$
When the potential of the cell is $$0.712 \mathrm{~V}$$ at $$298 \mathrm{~K}$$, the ratio $$\left[\mathrm{Fe}^{2+}\right] /\left[\mathrm{Fe}^{3+}\right]$$ is _____________. (Nearest integer)
Given : $$\mathrm{Fe}^{3+}+\mathrm{e}^{-}=\mathrm{Fe}^{2+}, \mathrm{E}^{\theta} \mathrm{Fe}^{3+}, \mathrm{Fe}^{2+} \mid \mathrm{Pt}=0.771$$
$$ \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V} $$
Explanation
Anode $${H_2} \to 2{H^ + } + 2{e^ - }$$
Cathode $$(F{e^{3 + }} + {e^ - } \to F{e^{2 + }}) \times 2$$
$${H_2} + 2F{e^{3 + }} \to 2{H^ + } + 2F{e^{2 + }}$$
$${E_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {\left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right)^2}$$
$$0.712 = 0.771 - 0.059\log {{F{e^{2 + }}} \over {F{e^{3 + }}}}$$
$$ - 0.059 = - 0.059\log {{F{e^{2 + }}} \over {F{e^{3 + }}}}$$
$${{[F{e^{2 + }}]} \over {[F{e^{3 + }}]}} = 10$$
Comments (0)
