Verified Solution • 2025–26 Session moderate

QJEEM25Ch4a9c: JEE Main 2025 Chemistry - Class 12 JEEM25Ch4a9c Solution: For the given cell The standard cell potential of the above…

For the given cell

Fe2+(aq)+Ag(aq)+→Fe3+(aq)+Ag(s)\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}_{(\mathrm{aq})}^{+} \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}_{(\mathrm{s})}

The standard cell potential of the above reaction is Given:

Ag++e−→Agamp;Eθ=xVFe2++2e−→Feamp;Eθ=yVFe3++3e−→Feamp;Eθ=zV\begin{array}{lr}\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag} & \mathrm{E}^\theta=\mathrm{xV} \\\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^\theta=\mathrm{yV} \\\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^\theta=\mathrm{zV}\end{array}

Multiple Choice Options

  • A x+2y−3zx+2 y-3 z
  • B x+2yx+2 y
  • C y−2xy-2 x
  • D x+y−zx+y-z

Step-by-Step Verified Solution

Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s)\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})

JEE Main 2025 (Online) 24th January Morning Shift Chemistry - Electrochemistry Question 3 English Explanation

amp;ΔG30=ΔG10+ΔG20amp;−3 F(−z)=−2 F(−y)+ΔG20amp;ΔG20=3Fz−2Fy\begin{aligned}& \Delta \mathrm{G}_3^0=\Delta \mathrm{G}_1^0+\Delta \mathrm{G}_2^0 \\& -3 \mathrm{~F}(-\mathrm{z})=-2 \mathrm{~F}(-\mathrm{y})+\Delta \mathrm{G}_2^0 \\& \Delta \mathrm{G}_2^0=3 \mathrm{Fz}-2 \mathrm{Fy}\end{aligned}

Also ΔG20=−nFEFe+2/Fe+30\Delta \mathrm{G}_2^0=-\mathrm{nFE}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0

amp;3Fz−2Fy=−1 F(EFe+2/Fe+30)amp;EFe+2/Fe+30=2y−3z\begin{aligned}& 3 \mathrm{Fz}-2 \mathrm{Fy}=-1 \mathrm{~F}\left(\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0\right) \\& \mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0=2 \mathrm{y}-3 \mathrm{z}\end{aligned}

ECell 0\mathrm{E}_{\text {Cell }}^0 for reaction will be

amp;EAg+/Ag00+EFe+2/Fe+30amp;=x+2y−3z\begin{aligned}& \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}^0}^0+\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0 \\& =\mathrm{x}+2 \mathrm{y}-3 \mathrm{z}\end{aligned}

Option (2)

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