Question

Give that at$\mathbf{2}\mathbf{5}^{\mathbf{o}}\mathbf{C}$, for

$Cr^{3 +}(aq) + e^{-} \rightarrow Cr^{2 +}(aq),E^{0} = - 0.424V$

$Cr^{2 +}(aq) + 2e^{-} \rightarrow Cr(s),E^{0} = - 0.900V$

find$E^{0}$at $25^{0}C$for $Cr^{3 +}(aq) + 3e \rightarrow Cr(s)$

• A. – 0.741V
• B.
  • 0.741 V
• C. – 1.324 V
• D. – 0.476 V

Answer 1

Answer: (A)

– 0.741V

Answer 2

Adding the first two reactions, we get the third equation and using the free energy concept, we have

$\Delta G_{1}^{0} + \Delta G_{2}^{0} = \Delta G_{3}^{0}$; $- n_{1}E_{1}^{0}F - n_{2}E_{2}^{0}F = - n_{3}E_{3}^{0}F$

(n = number of electron involved)

$\mathbf{E}_{\mathbf{3}}^{\mathbf{0}}\mathbf{=}\frac{\mathbf{n}_{\mathbf{1}}\mathbf{E}_{\mathbf{1}}^{\mathbf{0}}\mathbf{+}\mathbf{n}_{\mathbf{2}}\mathbf{E}_{\mathbf{2}}^{\mathbf{0}}}{\mathbf{n}_{\mathbf{3}}}\mathbf{=}\frac{\mathbf{1}\mathbf{\times}\mathbf{(}\mathbf{-}\mathbf{0.424) + 2}\mathbf{\times}\mathbf{(}\mathbf{-}\mathbf{0.900)}}{\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{0.741V}$