Verified Solution • 2025–26 Session moderate

QJEEM23Chc110: JEE Main 2023 Chemistry - Class 12 JEEM23Chc110 Solution: Consider the cell When the potential of the cell is at , the…

Consider the cell

Pt(s)∣H2( g,1 atm)∣H+(aq,1M)∣∣Fe3+(aq),Fe2+(aq)∣Pt⁡(s)\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 V0.712 \mathrm{~V} at 298 K298 \mathrm{~K} , the ratio [Fe2+]/[Fe3+]\left[\mathrm{Fe}^{2+}\right] /\left[\mathrm{Fe}^{3+}\right] is _____________. (Nearest integer)

Given : Fe3++e−=Fe2+,EθFe3+,Fe2+∣Pt=0.771\mathrm{Fe}^{3+}+\mathrm{e}^{-}=\mathrm{Fe}^{2+}, \mathrm{E}^{\theta} \mathrm{Fe}^{3+}, \mathrm{Fe}^{2+} \mid \mathrm{Pt}=0.771

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

Step-by-Step Verified Solution

Anode H2→2H++2e−{H_2} \to 2{H^ + } + 2{e^ - }

Cathode (Fe3++e−→Fe2+)×2(F{e^{3 + }} + {e^ - } \to F{e^{2 + }}) \times 2

H2+2Fe3+→2H++2Fe2+{H_2} + 2F{e^{3 + }} \to 2{H^ + } + 2F{e^{2 + }}

Ecell=Ecello−0.0592log⁡(Fe2+Fe3+)2{E_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {\left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right)^2}

0.712=0.771−0.059log⁡Fe2+Fe3+0.712 = 0.771 - 0.059\log {{F{e^{2 + }}} \over {F{e^{3 + }}}}

−0.059=−0.059log⁡Fe2+Fe3+- 0.059 = - 0.059\log {{F{e^{2 + }}} \over {F{e^{3 + }}}}

[Fe2+][Fe3+]=10{{[F{e^{2 + }}]} \over {[F{e^{3 + }}]}} = 10

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