Exam

JEE Main

Year

2025

Date

24 January

Shift

Morning Shift

Subject

Chemistry

Chapter

Electrochemistry

JEE Main 2025 · 24 January · Morning Shift · Chemistry · Electrochemistry

For the given cell Fe^ 2+ _ (aq) +Ag^ + _ (aq) arrow Fe_ (aq) ^ 3+...

For the given cell $\mathrm{Fe}^{2+}{ }_{(\mathrm{aq})}+\mathrm{Ag}^{+}{ }_{(\mathrm{aq})} \rightarrow \mathrm{Fe}_{(\mathrm{aq})}^{3+}+\mathrm{Ag}_{(\mathrm{s})}$ The standard cell potential of the above reaction is Given: $$\begin{array}{lr} \mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag} & \mathrm{E}^\theta=\mathrm{xV} \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^\theta=\mathrm{yV} \\ \mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^\theta=\mathrm{zV} \end{array}$$

Step-by-step solution

$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$
$$\begin{aligned} & \Delta \mathrm{G}_3^0=\Delta \mathrm{G}_1^0+\Delta \mathrm{G}_2^0 \\ & -3 \mathrm{~F}(-\mathrm{z})=-2 \mathrm{~F}(-\mathrm{y})+\Delta \mathrm{G}_2{ }^0 \\ & \Delta \mathrm{G}_2^0=3 \mathrm{Fz}-2 \mathrm{Fy} \end{aligned}$$ Also $\Delta \mathrm{G}_2^0=-\mathrm{nFE}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{t^{43}}}^0$$\begin{aligned} & 3 \mathrm{Fz}-2 \mathrm{Fy}=-1 \mathrm{~F}\left(\mathrm{E}_{\mathrm{Fe}^{t^2} / \mathrm{Fe}^{43}}^0\right) \\ & \mathrm{E}_{\mathrm{Fe}^{2+2} / \mathrm{Fe}^{t^3}}^0=2 \mathrm{y}-3 \mathrm{z} \end{aligned}$$\mathrm{E}_{\text{Cell }}^0$ for reaction will be $$\begin{aligned} & \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0+\mathrm{E}_{\mathrm{Fe}^{t^2 / \mathrm{Fe}}}^{03} \\ & =\mathrm{x}+2 \mathrm{y}-3 \mathrm{z} \end{aligned}$$