Q30 / 90 · JEE Main 2021 (27 Jul Shift 2) · Chemistry

For the cell $Cu (s) |{Cu}^{2 +} (aq) (0.1 M) ‖ {Ag}^{+} (aq) (0.01 M)| Ag (s)$ the cell potential $E_{1} = 0.3095 V$ . For the cell $Cu (s) |{Cu}^{2 +} (aq) (0.01 M) ‖ {Ag}^{+} (aq) (0.001 M)| Ag (s)$ the cell potential $= x \times 10^{- 2} V .$ Find value of x (Round off the Nearest Integer).
[ Use $: \frac{2.303 RT}{F} = 0.059 J$ ]

Correct answer: 28

Solution

Cell reaction is :

$Cu (s) + 2 {Ag}^{+} (aq) \to {Cu}^{2 +} (aq) + 2 Ag (s)$

Now, $E_{cell} = E_{Cell}^{o} - \frac{0.059}{2} \log \frac{[{Cu}^{2 +}]}{{[{Ag}^{+}]}^{2}}$

$E_{1} = 0.3095 = E_{Cell}^{o} - \frac{0.059}{2} \cdot \log \frac{0.1}{(0.01)^{2}}$ ....... $(1)$

$E_{2} = E_{Cell}^{o} - \frac{0.059}{2} \cdot \log \frac{0.01}{(0.001)^{2}} . . . . . . . . . . (2)$

From $(1)$ and $(2), E_{2} = 0.28 V = 28 \times 10^{- 2} V$

Hence, the correct answer is 28.