The EM3 + / M2 +o values for Cr, Mn, Fe and Co are -0.41, +1.57, +0.77 and +1.97 V respectively. For which one of following metal the change in oxidation state from +2 to +3 easiest?

Question

The EM3 + / M2 +o values for Cr, Mn, Fe and Co are -0.41, +1.57, +0.77 and +1.97 V respectively. For which one of following metal the change in oxidation state from +2 to +3 easiest? (A) Co (B) Mn (C) Fe (D) Cr

Tardigrade Admin · Accepted Answer

Here ER Po of Cr = - 0.41 so its EO Po is 0.41. As here EO Po for Cr2 +/Cr3 + is maximum +0.41 V so Cr2 + will be easily oxidised to Cr3 + . For rest of the metals EO Po will be negative so they have less tendency to be oxidized then Cr here.

Q. The $E_{M^{3 +} / M^{2 +}}^{o}$ values for Cr, Mn, Fe and Co are -0.41, +1.57, +0.77 and +1.97 V respectively. For which one of following metal the change in oxidation state from +2 to +3 easiest?

Co

Mn

Fe

Cr

Here $E_{RP}^{o}$ of Cr = - 0.41 so its $E_{OP}^{o}$ is 0.41.
As here $E_{OP}^{o}$ for $C r^{2 +} / C r^{3 +}$ is maximum +0.41 V so $C r^{2 +}$ will be easily oxidised to $C r^{3 +}$ .
For rest of the metals $E_{OP}^{o}$ will be negative so they have less tendency to be oxidized then Cr here.

Q. The EM3+/M2+o​ values for Cr, Mn, Fe and Co are -0.41, +1.57, +0.77 and +1.97 V respectively. For which one of following metal the change in oxidation state from +2 to +3 easiest?

Co

Mn

Fe

Cr

Here ERPo​ of Cr = - 0.41 so its EOPo​ is 0.41. As here EOPo​ for Cr2+/Cr3+ is maximum +0.41 V so Cr2+ will be easily oxidised to Cr3+ . For rest of the metals EOPo​ will be negative so they have less tendency to be oxidized then Cr here.

Q. The $E_{M^{3 +} / M^{2 +}}^{o}$ values for Cr, Mn, Fe and Co are -0.41, +1.57, +0.77 and +1.97 V respectively. For which one of following metal the change in oxidation state from +2 to +3 easiest?

Here $E_{RP}^{o}$ of Cr = - 0.41 so its $E_{OP}^{o}$ is 0.41.
As here $E_{OP}^{o}$ for $C r^{2 +} / C r^{3 +}$ is maximum +0.41 V so $C r^{2 +}$ will be easily oxidised to $C r^{3 +}$ .
For rest of the metals $E_{OP}^{o}$ will be negative so they have less tendency to be oxidized then Cr here.