The cell potential for the given cell at 298 K $\mathrm{Pt}\left|\mathrm{H}_{2}(\mathrm{g}, 1\ \mathrm{bar})\right| \mathrm{H}^{+}(\mathrm{aq}) \| \mathrm{Cu}^{2+}(\mathrm{aq}) \mid \mathrm{Cu}(\mathrm{s})$ is 0.31 V. The pH of the acidic solution is found to be 3, whereas the concentration of $\mathrm{Cu}^{2+}$ is $10^{-x}\ \mathrm{M}$. The value of $x$ is \_\_\_\_ . (Given: $E^{\ominus}_{\mathrm{Cu}^{2+}/\mathrm{Cu}}=0.34\ \mathrm{V}$ and $\dfrac{2.303\,RT}{F}=0.06\ \mathrm{V}$)

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Redox Reactions and Electrochemistry

The cell potential for the given cell at 298 K

Pt ∣ H_{2} (g, 1 bar) ∣ H^{+} (aq) ∥ Cu^{2 +} (aq) ∣ Cu (s)

is 0.31 V. The pH of the acidic solution is found to be 3, whereas the concentration of

Cu^{2 +}

1 0^{- x} M

. The value of

x

is ____ .
(Given:

E_{Cu^{2 +} / Cu}^{⊖} = 0.34 V

and

\frac{2.303 R T}{F} = 0.06 V

)

Answer: 7

The cell reaction is:

H_{2} (g) + Cu^{2 +} (aq) \to 2 H^{+} (aq) + Cu (s)

Using the Nernst equation:

E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.059}{n} lo g \frac{[ H ^{+} ] ^{2}}{[ Cu ^{2 +} ]}

Given:

$E_{ce l l} = 0.31$ V
$E_{ce l l}^{\circ} = E_{Cu^{2 +} / Cu}^{\circ} - E_{H^{+} / H_{2}}^{\circ} = 0.34 - 0 = 0.34$ V
pH = 3, so $[H^{+}] = 1 0^{- 3}$ M
$n = 2$

Substituting:

0.31 = 0.34 - \frac{0.06}{2} lo g \frac{( 1 0 ^{- 3} ) ^{2}}{[ Cu ^{2 +} ]}

0.31 = 0.34 - 0.03 lo g \frac{1 0 ^{- 6}}{[ Cu ^{2 +} ]}

- 0.03 = - 0.03 lo g \frac{1 0 ^{- 6}}{[ Cu ^{2 +} ]}

1 = lo g \frac{1 0 ^{- 6}}{[ Cu ^{2 +} ]}

10 = \frac{1 0 ^{- 6}}{[ Cu ^{2 +} ]}

[Cu^{2 +}] = 1 0^{- 7}

M

Therefore,

x = 7

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Integer answer