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The reduction potential of a half cell consisting of a Pt electrode immersed in $2.0 \mathrm{M} \mathrm{Fe}^{2+}$ and $0.02 \mathrm{M} \mathrm{Fe}^{3+}$ solution (in V) is
Given $\left(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059, \mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^0=0.771 \mathrm{~V}\right)$
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Given $\left(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059, \mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^0=0.771 \mathrm{~V}\right)$
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Verified Answer
The correct answer is:
0.653
$\mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}$
Using Nernst equation,
$\mathrm{E}_{\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)}=\mathrm{E}_{\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)} \frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]}$
$\begin{aligned} & \text { Here, } \mathrm{n}=1,\left[\mathrm{Fe}^{2+}\right]=2 \mathrm{M},\left[\mathrm{Fe}^{3+}\right]=0.02 \mathrm{M}, \\ & \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059, \mathrm{E}^{\circ}=0.771 \mathrm{~V}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right) \\ & \mathrm{E}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=0.771-0.059 \log \frac{2}{0.02} \\ & \mathrm{E}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=0.653 \mathrm{~V}\end{aligned}$
Using Nernst equation,
$\mathrm{E}_{\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)}=\mathrm{E}_{\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)} \frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]}$
$\begin{aligned} & \text { Here, } \mathrm{n}=1,\left[\mathrm{Fe}^{2+}\right]=2 \mathrm{M},\left[\mathrm{Fe}^{3+}\right]=0.02 \mathrm{M}, \\ & \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059, \mathrm{E}^{\circ}=0.771 \mathrm{~V}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right) \\ & \mathrm{E}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=0.771-0.059 \log \frac{2}{0.02} \\ & \mathrm{E}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=0.653 \mathrm{~V}\end{aligned}$
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