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The standard reduction potential \(\mathrm{E}^{\circ}\) for half reations are
\(\begin{array}{ll}
\mathrm{Zn}=\mathrm{Zn}^{+2}+\mathrm{Ze} & \mathrm{E}^{\circ}=+0.76 \mathrm{~V} \\
\mathrm{Fe}=\mathrm{Fe}^{+2}+\mathrm{Ze} & \mathrm{E}^{\circ}=+0.41 \mathrm{~V}
\end{array}\)
The EMF of hte cell reaction \(\mathrm{Fe}^{+2}+\mathrm{Zn}=\mathrm{Zn}^{+2}+\mathrm{Fe}\) is
Options:
\(\begin{array}{ll}
\mathrm{Zn}=\mathrm{Zn}^{+2}+\mathrm{Ze} & \mathrm{E}^{\circ}=+0.76 \mathrm{~V} \\
\mathrm{Fe}=\mathrm{Fe}^{+2}+\mathrm{Ze} & \mathrm{E}^{\circ}=+0.41 \mathrm{~V}
\end{array}\)
The EMF of hte cell reaction \(\mathrm{Fe}^{+2}+\mathrm{Zn}=\mathrm{Zn}^{+2}+\mathrm{Fe}\) is
Solution:
2314 Upvotes
Verified Answer
The correct answer is:
+0.35
\(\begin{aligned}
& \text {Hints: } \mathrm{E}_{\text {cell }}=\mathrm{E}_{\operatorname{Anu}([\mathrm{Qp})}^{\circ}-\mathrm{E}_{\text {cathode }(\mathrm{op})}^{\circ} \\
& =0.76-0.41 \\
& =+0.35 \mathrm{~V}
\end{aligned}\)
& \text {Hints: } \mathrm{E}_{\text {cell }}=\mathrm{E}_{\operatorname{Anu}([\mathrm{Qp})}^{\circ}-\mathrm{E}_{\text {cathode }(\mathrm{op})}^{\circ} \\
& =0.76-0.41 \\
& =+0.35 \mathrm{~V}
\end{aligned}\)
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