QJEEM25Chf9e2: JEE Main 2025 Chemistry - Class 12 JEEM25Chf9e2 Solution: FeO 4 2 − → + 2.0   V Fe 3 + → 0.8   V Fe 2 + → − 0.5 …

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QJEEM25Chf9e2: JEE Main 2025 Chemistry - Class 12 JEEM25Chf9e2 Solution: FeO 4 2 − → + 2.0   V Fe 3 + → 0.8   V Fe 2 + → − 0.5 …

${FeO}_{4}^{2 -} \overset{+ 2.0 V}{\to} {Fe}^{3 +} \overset{0.8 V}{\to} {Fe}^{2 +} \overset{- 0.5 V}{\to} {Fe}^{0}$

In the above diagram, the standard electrode potentials are given in volts (over the arrow).

The value of $\mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{2+}}$ is :

Multiple Choice Options

A 1.2 V
B 2.1 V
C 1.4 V
D 1.7 V

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Accepted Answer

JEE Main 2025 (Online) 23rd January Morning Shift Chemistry - Electrochemistry Question 5 English Explanation

$\begin{aligned} Δ G_{4}^{o} = Δ G_{1}^{o} + Δ G_{2}^{o} \\ \Rightarrow & - n_{4} {FE}_{4}^{o} = - n_{1} {FE}_{1}^{0} - n_{2} {FE}_{2}^{o} \\ \Rightarrow & + 4 E_{4}^{o} = 3 \times 2 + (1 \times 0.8) \\ \Rightarrow & E_{4}^{o} = \frac{6.8}{4} V \\ \Rightarrow & E_{4}^{o} = 1.7 V \end{aligned}$

QJEEM25Chf9e2: JEE Main 2025 Chemistry - Class 12 JEEM25Chf9e2 Solution: FeO 4 2 − → + 2.0 &nbsp; V Fe 3 + → 0.8 &nbsp; V Fe 2 + → − 0.5 …

Multiple Choice Options

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QJEEM25Chf9e2: JEE Main 2025 Chemistry - Class 12 JEEM25Chf9e2 Solution: FeO 4 2 − → + 2.0 V Fe 3 + → 0.8 V Fe 2 + → − 0.5 …