The hydrogen electrode is dipped in a solution of $\mathrm{pH}=3$ at $25^{\circ} \mathrm{C}$. The potential of the electrode will be - $\_\_\_\_$ $\times 10^{-2} \mathrm{~V}$. $$\left(\frac{2.303 \mathrm{RT}}{\mathrm{~F}}=0.059 \mathrm{~V}\right)$$

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Redox Reactions and Electrochemistry

The hydrogen electrode is dipped in a solution of

pH = 3

2 5^{\circ} C

. The potential of the electrode will be -

____

\times 1 0^{- 2} V

(\frac{2.303 RT}{F} = 0.059 V)

Answer: 18

For the hydrogen electrode,

2 H^{+} + 2 e^{-} \to H_{2}

, the Nernst equation gives

E = E^{\circ} - \frac{0.059}{2} lo g \frac{P _{H_{2}}}{[ H ^{+} ] ^{2}}

.

With

E^{\circ} = 0

P_{H_{2}} = 1

atm and

p H = 3

[H^{+}] = 1 0^{- 3}

E = - \frac{0.059}{2} lo g (1 0^{6}) = - \frac{0.059}{2} \times 6 = - 0.177

= - 17.7 \times 1 0^{- 2}

V.

The magnitude to the nearest integer is 18.

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Integer answer