Using the data given below find out the strongest reducing agent. E° Cr 2 O 4 / Cr 3+=1.33 V , E° Cl 2 / Cl -=1.36 V, E° MnO 4- / Mn 2+=1.51 V , E° Cr 3+ / Cr =-0.74 V

Question

Using the data given below find out the strongest reducing agent. E° Cr 2 O 4 / Cr 3+=1.33 V , E° Cl 2 / Cl -=1.36 V, E° MnO 4- / Mn 2+=1.51 V , E° Cr 3+ / Cr =-0.74 V (A) Cl- (B) Mn2+ (C) Cr (D) Cr3+

Tardigrade Admin · Accepted Answer

Strongest reducing agent is one which is oxidised most easily i.e., which has highest oxidation potential. Cr / Cr 3+ has highest oxidation potential i.e., (+0.74) so, it is oxidised most easily and therefore, it is the strongest reducing agent.

Q. Using the data given below find out the strongest reducing agent.
$E^{\circ} C r_{2} O_{4} / C r^{3 +} = 1.33 V, E^{\circ} C l_{2} / C l^{-} = 1.36 V$ ,
$E^{\circ} M n O_{4}^{-} / M n^{2 +} = 1.51 V, E^{\circ} C r^{3 +} / C r = - 0.74 V$

$C l^{-}$

$M n^{2} +$

$C r$

$C r^{3 +}$

Strongest reducing agent is one which is oxidised most easily i.e., which has highest oxidation potential. $C r / C r^{3 +}$ has highest oxidation potential i.e., $(+ 0.74)$ so, it is oxidised most easily and therefore, it is the strongest reducing agent.

Q. Using the data given below find out the strongest reducing agent. E∘Cr2​O4​/Cr3+=1.33V,E∘Cl2​/Cl−=1.36V, E∘MnO4−​/Mn2+=1.51V,E∘Cr3+/Cr=−0.74V

Cl−

Mn2+

Cr

Cr3+