From the following E° values of half-cells; i. A+e arrow A-;E° =-0.24 V ii. B-+e arrow B2 -:E° =+1.25 V iii. C-+2e arrow C3;E0=-1.25 V iv. D+2e arrow D2 -;E0=+0.68 V If you were to construct a cell using combination of two half-cells from above that gives the largest cell potential, then value of the largest cell potential would be (x/2) V. Value of x will be.

Question

From the following E° values of half-cells; i. A+e arrow A-;E° =-0.24 V ii. B-+e arrow B2 -:E° =+1.25 V iii. C-+2e arrow C3;E0=-1.25 V iv. D+2e arrow D2 -;E0=+0.68 V If you were to construct a cell using combination of two half-cells from above that gives the largest cell potential, then value of the largest cell potential would be (x/2) V. Value of x will be. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Larger the difference between the electrode potentials, greater is the cell potential. Hence, combination of half-cells

(ii)

and

(iii)

will give the largest cell potential.

E_{c ll}^{\circ} = E_{c a l s a d e}^{\circ} - E_{r mak}^{\circ} = 1.25 - (- 1.25)

= + 2.50 V = \frac{5}{2} V

Larger the difference between the electrode potentials, greater is the cell potential. Hence, combination of half-cells (ii) and (iii) will give the largest cell potential. Ecll∘​=Ecalsade∘​−Ermak∘​=1.25−(−1.25) =+2.50V=25​V

Larger the difference between the electrode potentials, greater is the cell potential. Hence, combination of half-cells $(ii)$ and $(iii)$ will give the largest cell potential.
$E_{c ll}^{\circ} = E_{c a l s a d e}^{\circ} - E_{r mak}^{\circ} = 1.25 - (- 1.25)$
$= + 2.50 V = \frac{5}{2} V$