JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
\(\mathrm{FeO}_4^{2-} \xrightarrow{+2.0 \mathrm{v}} \mathrm{Fe}^{3+} \xrightarrow{0.8 \mathrm{v}} \mathrm{Fe}^{2+} \xrightarrow{-0.5 \mathrm{v}} \mathrm{Fe}^0\)
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 _______.
- A 2.1 V
- B 1 .7 V
- C 1 .4 V
- D 1 .2 V
Answer & Solution
Correct Answer
(B) 1 .7 V
Step-by-step Solution
Detailed explanation
\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…
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