For the cell \(\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(\mathrm{aq})(0.1 \mathrm{M})\right|\left|\mathrm{Ag}^{+}(\mathrm{aq})(0.01 \mathrm{M})\right| \mathrm{Ag}(\mathrm{s})\) The cell potential \(\mathrm{E}_{1}=0.3095 \mathrm{~V}\) For the cell \(\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(\mathrm{aq})(0.01 \mathrm{M})\right|\left|\mathrm{Ag}^{+}(\mathrm{aq})(0.001 \mathrm{M})\right| \mathrm{Ag}(\mathrm{s})\) The cell potential = \_\_\_\_ \(\times 10^{-2} \mathrm{~V}\). (Round off to the Nearest Integer). [Use : \(\frac{2.303 R T}{F}=0.059\) ]

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Redox Reactions and Electrochemistry

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 = ____

\times 1 0^{- 2} V

. (Round off to the Nearest Integer).
[Use :

\frac{2.303 R T}{F} = 0.059

]

Answer: 28

The cell reaction is

Cu + 2 A g^{+} \to C u^{2 +} + 2 Ag

, so

n = 2

and

E = E^{\circ} - \frac{0.059}{2} lo g \frac{[ C u ^{2 +} ]}{[ A g ^{+} ] ^{2}}

For the first cell the reaction quotient is

0.1/ (0.01)^{2} = 1 0^{3}

, so

0.3095 = E^{\circ} - \frac{0.059}{2} (3) = E^{\circ} - 0.0885 \Rightarrow E^{\circ} = 0.398 V

For the second cell the quotient is

0.01/ (0.001)^{2} = 1 0^{4}

, so

E = 0.398 - \frac{0.059}{2} (4) = 0.398 - 0.118 = 0.280 V

The potential of the second cell is

28 \times 1 0^{- 2}

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer