Search any question & find its solution
Question:
Answered & Verified by Expert
Consider the following $\mathrm{E}^{\circ}$ values
$$
\begin{aligned}
& \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=0.77 \mathrm{~V} \\
& \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}=-0.14 \mathrm{~V}
\end{aligned}
$$
Under standard conditions the potential for the reaction
$\mathrm{Sn}(\mathrm{s})+2 \mathrm{Fe}^{3+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})$ is
Options:
$$
\begin{aligned}
& \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=0.77 \mathrm{~V} \\
& \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}=-0.14 \mathrm{~V}
\end{aligned}
$$
Under standard conditions the potential for the reaction
$\mathrm{Sn}(\mathrm{s})+2 \mathrm{Fe}^{3+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})$ is
Solution:
2453 Upvotes
Verified Answer
The correct answer is:
$0.91 \mathrm{~V}$
$0.91 \mathrm{~V}$
$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\mathrm{RHS}}^{\circ}-\mathrm{E}_{\mathrm{LHS}}^{\circ}$
$=(0.77)-(-0.14)$
$=0.91 \mathrm{~V}$
$=(0.77)-(-0.14)$
$=0.91 \mathrm{~V}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.