A cell is formed by combining given electrodes with \( Zn^{2+}| Zn(s) \) (E° = -0.76 V). Arrange the following in decreasing order of E° cell.
\( Zn(s)|Zn^{2+} \) || Given electrode
(A) \( Ag^+| Ag(s)\,\,\, E° = 0.80 V \)
(B) \( Na^+| Na(s)\,\,\, E° = -2.71 V \)
(C) \( Fe^{2+}| Fe(s)\,\,\, E° = -0.44 V \)
(D) \( Cu^+| Cu(s)\,\,\, E° = 0.52 V \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (D), (C), (B)
The standard emf of a cell is the standard electrode potential of the right-hand electrode (cathode) minus that of the left-hand electrode (anode):
\( \displaystyle E^\circ_{cell} = E^\circ_{right} - E^\circ_{left} \)
In every cell here the zinc electrode is on the left, \( E^\circ_{left} = -0.76 \) V, so \( E^\circ_{cell} = E^\circ_{given} + 0.76 \) V:
(A) \( \ce{Ag+ | Ag} \): \( 0.80 - (-0.76) = 1.56 \) V
(B) \( \ce{Na+ | Na} \): \( -2.71 - (-0.76) = -1.95 \) V
(C) \( \ce{Fe^2+ | Fe} \): \( -0.44 - (-0.76) = 0.32 \) V
(D) \( \ce{Cu+ | Cu} \): \( 0.52 - (-0.76) = 1.28 \) V
The negative value for sodium means that cell would not work as written: sodium, with the lower potential, would itself be the anode.
Shortcut: The anode is the same in all four cells, so \( E^\circ_{cell} \) follows the order of the given electrode potentials: 0.80 > 0.52 > −0.44 > −2.71, that is (A), (D), (C), (B), with no subtraction needed.