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On the basis of standard electrode potential of redox couples given below, find out which of the following is the strongest oxidising agent. $\left(E^{\circ}\right.$ values $: \mathrm{Fe}^{3+}\left|\mathrm{Fe}^{2+}=+0.77 \mathrm{~V} ; \mathrm{I}_{2(s)}\right| \mathrm{I}^{-}=$ $\left.+0.54 \mathrm{~V} ; \mathrm{Cu}^{2+}\left|\mathrm{Cu}=+0.34 \mathrm{~V} ; \mathrm{Ag}^{+}\right| \mathrm{Ag}=+0.80 \mathrm{~V}\right)$
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Verified Answer
The correct answer is:
$\mathrm{Ag}^{+}$
Higher the electrode potential, better is the oxidising agent. Since, the electrode
potential decreases in the order :
$\mathrm{Ag}^{+}\left|\mathrm{Ag}(+0.80 \mathrm{~V})>\mathrm{Fe}^{3+}\right| \mathrm{Fe}^{2+}(+0.77 \mathrm{~V})>$
$\mathrm{I}_{2(s)}\left|\mathrm{I}^{-}(+0.54 \mathrm{~V})>\mathrm{Cu}^{2+}\right| \mathrm{Cu}(+0.34 \mathrm{~V})$
Hence, $\mathrm{Ag}^{+}$is the strongest oxidising agent.
potential decreases in the order :
$\mathrm{Ag}^{+}\left|\mathrm{Ag}(+0.80 \mathrm{~V})>\mathrm{Fe}^{3+}\right| \mathrm{Fe}^{2+}(+0.77 \mathrm{~V})>$
$\mathrm{I}_{2(s)}\left|\mathrm{I}^{-}(+0.54 \mathrm{~V})>\mathrm{Cu}^{2+}\right| \mathrm{Cu}(+0.34 \mathrm{~V})$
Hence, $\mathrm{Ag}^{+}$is the strongest oxidising agent.
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