$Li$ occupies a higher position in the electrochemical series of metals as compared to $Cu$ since:
- Athe standard reduction potential of $Li^{+} / Li$ is lower than that of $Cu^{2+} / Cu$
- Bthe standard reduction potential of $Cu^{2+} / Cu$ is lower than that of $Li^{+} / Li$
- Cthe standard oxidation potential of $Li / Li^{+}$ is lower than that of $Cu / Cu^{2+}$
- D$Li$ is smaller in size as compared to $Cu$
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Using the data given below,find out the strongest reducing agent:
$E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{Cl_2/Cl^{-}} = 1.36 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}$
$E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{Cl_2/Cl^{-}} = 1.36 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}$
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View SolutionArrange the following metals in the order of their decreasing standard electrode potential: $Mg, K, Ba, Ca$.
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