MCQ
Using the data given in find out in which option the order of reducing power is correct.
$\text{E}^\ominus_{\text{cr}_2\text{O}^{2-}_7/\text{cr}^{3+}}=1.33\text{V}\text{E}^\ominus_{\text{Cl}_2/\text{Cl}^-}=1.36\text{V}$
$\text{E}^\ominus_{\text{MnO}^-_4/\text{Mn}^{2+}}=1.51\text{V}\text{E}^\ominus_{\text{Cr}^{3+}/\text{Cr}}=-0.74\text{V}$
$\text{E}^\ominus_{\text{cr}_2\text{O}^{2-}_7/\text{cr}^{3+}}=1.33\text{V}\text{E}^\ominus_{\text{Cl}_2/\text{Cl}^-}=1.36\text{V}$
$\text{E}^\ominus_{\text{MnO}^-_4/\text{Mn}^{2+}}=1.51\text{V}\text{E}^\ominus_{\text{Cr}^{3+}/\text{Cr}}=-0.74\text{V}$
- A$\text{Cr}^{3+}<\text{Cl}^-<\text{Mn}^{2+}<\text{Cr}$
- ✓$\text{Mn}^{2+}<\text{Cl}^-<\text{Cr}^{3+}<\text{Cr}$
- C$\text{Cr}^{3+}<\text{Cl}^{-}<\text{Cr}_2\text{O}^{2-}_7<\text{MnO}_4^{-}$
- D$\text{Mn}^{2+}<\text{Cr}^{3+}<\text{Cl}^{-}<\text{Cr}$
