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:

(A), (D), (C), (B)

The correct answer is Option (1) → (A), (D), (C), (B)

Given cell:
Zn(s) | Zn²⁺ || Given electrode

Standard cell potential:

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$

Here, Zn²⁺/Zn is the anode with

$E^\circ_{\text{anode}} = -0.76\ \text{V}$

So for each case:

$E^\circ_{\text{cell}} = E^\circ_{\text{given}} - (-0.76) = E^\circ_{\text{given}} + 0.76$

Now calculate:

• (A) Ag⁺/Ag: $0.80 + 0.76 = 1.56\ \text{V}$
• (D) Cu⁺/Cu: $0.52 + 0.76 = 1.28\ \text{V}$
• (C) Fe²⁺/Fe: $-0.44 + 0.76 = 0.32\ \text{V}$
• (B) Na⁺/Na: $-2.71 + 0.76 = -1.95\ \text{V}$

Decreasing order of $E^\circ_{\text{cell}}$​: $(A) > (D) > (C) > (B)$