JEE MAIN - Chemistry (2025 - 23rd January Morning Shift - No. 5)
$$
\mathrm{FeO}_4^{2-} \xrightarrow{+2.0 \mathrm{~V}} \mathrm{Fe}^{3+} \xrightarrow{0.8 \mathrm{~V}} \mathrm{Fe}^{2+} \xrightarrow{-0.5 \mathrm{~V}} \mathrm{Fe}^0
$$
In the above diagram, the standard electrode potentials are given in volts (over the arrow).
The value of $\mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{2+}}$ is :
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 :
1.2 V
2.1 V
1.4 V
1.7 V
Explanation
_23rd_January_Morning_Shift_en_5_1.png)
$$\begin{aligned} & \Delta \mathrm{G}_4^{\mathrm{o}}=\Delta \mathrm{G}_1^{\mathrm{o}}+\Delta \mathrm{G}_2^{\mathrm{o}} \\ \Rightarrow & -\mathrm{n}_4 \mathrm{FE}_4^{\mathrm{o}}=-\mathrm{n}_1 \mathrm{FE}_1^0-\mathrm{n}_2 \mathrm{FE}_2^{\mathrm{o}} \\ \Rightarrow & +4 \mathrm{E}_4^{\mathrm{o}}=3 \times 2+(1 \times 0.8) \\ \Rightarrow & \mathrm{E}_4^{\mathrm{o}}=\frac{6.8}{4} \mathrm{~V} \\ \Rightarrow & \mathrm{E}_4^{\mathrm{o}}=1.7 \mathrm{~V} \end{aligned}$$
Comments (0)
