Medium INTEGER +4 / -1 PYQ · JEE Mains 2023

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 : $$\mathrm{Fe}^{3+}+\mathrm{e}^{-}=\mathrm{Fe}^{2+}, \mathrm{E}^{\theta} \mathrm{Fe}^{3+}, \mathrm{Fe}^{2+} \mid \mathrm{Pt}=0.771$$

$\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}$

Answer (integer) 10

Solution

<p>Anode ${H_2} \to 2{H^ + } + 2{e^ - }$</p> <p>Cathode $(F{e^{3 + }} + {e^ - } \to F{e^{2 + }}) \times 2$</p> <p>${H_2} + 2F{e^{3 + }} \to 2{H^ + } + 2F{e^{2 + }}$</p> <p>$${E_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {\left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right)^2}$$</p> <p>$0.712 = 0.771 - 0.059\log {{F{e^{2 + }}} \over {F{e^{3 + }}}}$</p> <p>$- 0.059 = - 0.059\log {{F{e^{2 + }}} \over {F{e^{3 + }}}}$</p> <p>${{[F{e^{2 + }}]} \over {[F{e^{3 + }}]}} = 10$</p>

About this question

Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells

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