Exam

JEE Main

Year

2025

Date

23 January

Shift

Morning Shift

Subject

Chemistry

Chapter

Electrochemistry

JEE Main 2025 · 23 January · Morning Shift · Chemistry · Electrochemistry

FeO_4^ 2- +2.0 v Fe^ 3+ 0.8 v Fe^ 2+ -0.5 v Fe^0 In the above...

$\mathrm{FeO}_4^{2-} \xrightarrow{+2.0 \mathrm{v}} \mathrm{Fe}^{3+} \xrightarrow{0.8 \mathrm{v}} \mathrm{Fe}^{2+} \xrightarrow{-0.5 \mathrm{v}} \mathrm{Fe}^0$ In the above diagram, the standard electrode potentials are given in volts (over the arrow). The value of $\mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{2+}}^{\mathrm{O}}$ is

Step-by-step solution

$$\begin{aligned} & \Delta \mathrm{G}_4^{\mathrm{o}}=\Delta \mathrm{G}_1^{\mathrm{o}}+\Delta \mathrm{G}_2^{\mathrm{o}} \\ \Rightarrow & -\mathrm{n}_4 \mathrm{FE}_4^{\mathrm{o}}=-\mathrm{n}_1 \mathrm{FE}_1^0-\mathrm{n}_2 \mathrm{FE}_2^{\mathrm{o}} \\ \Rightarrow & +4 \mathrm{E}_4^{\mathrm{o}}=3 \times 2+(1 \times 0.8) \\ \Rightarrow & \mathrm{E}_4^{\mathrm{o}}=\frac{6.8}{4} \mathrm{~V} \\ \Rightarrow & \mathrm{E}_4^{\mathrm{o}}=1.7 \mathrm{~V}\end{aligned}$$