CBSE 2024 · Region 1 · Set 1 · Q17 · 2 marks
Calculate the potential of Iron electrode in which the concentration of $\displaystyle \mathrm{Fe}^{2+}$ ion is $\displaystyle 0.01$ M. ( $\displaystyle \mathrm{E}^{0} \mathrm{Fe}^{2+} / \mathrm{Fe}=-0.45 \mathrm{~V}$ at $\displaystyle 298$ K ) [Given : $\displaystyle \log 10=1$ ]
Marking-scheme solution
\[\begin{aligned}
& \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}^{2}}=\mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}^{-\frac{0.059}{2}} \log \frac{1}{\left[F e^{2+}\right]}}^{\mathrm{O}} \\
& =-0.45 \mathrm{~V}-\frac{0.059}{2} \log \frac{1}{0.01} \\
& =-0.45 \mathrm{~V}-0.059 \mathrm{~V} \\
& =-0.509 \mathrm{~V}
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.