The hydrogen electrode is dipped in a solution of pH = 3 at 25^ C. The potential of the electrode will be - x 10^-2 V.

Given: hydrogen electrode, pH = 3, temperature 25^ C. Find: the value of x in the potential expression -x 10^-2 V. For the hydrogen electrode, the reaction is: H2(g) 2 H^+(aq) + 2e^- Using the Nernst equation: E = E^ - 2.303RTnF ( [ H^+]^2P H2) For the standard hydrogen electrode, E^ = 0, P H2 = 1 atm, and at pH = 3, [ H^+] = 10^-3 So the equation simplifies to: E = -0.059 pH Substituting pH = 3: E = -0.059 3 V E = -0.177 V Hence, E = -17.7 10^-2 V Therefore, x = 17.7 18, so the answer is 18. Given: E^0 = 0 for the standard hydrogen electrode and pH = 3, so [ H^+] = 10^-3 M Find: the numerical value of x. Using the relation stated in the solution: E = E^0 - 0.059 (10^3) E = 0 - 0.059 3 E = -0.177 V Rewriting in the required form: -0.177 V = -17.7 10^-2 V Thus the required value is 18 after rounding to the nearest integer.

The hydrogen electrode is dipped in a solution of pH = 3 at 25^ C… — Solution | JEE Chemistry

NVAEasyJEE 2024Nernst Equation

The hydrogen electrode is dipped in a solution of pH = $3$ at $25^\circ \text{C}$ . The potential of the electrode will be $- x \times 10^{-2} \, \text{V}$ .

Answer

Correct answer:18

Step-by-step solution

Standard Method

Given: hydrogen electrode, $\text{pH} = 3$ , temperature $25^\circ \text{C}$ .

Find: the value of $x$ in the potential expression $-x \times 10^{-2} \, \text{V}$ .

For the hydrogen electrode, the reaction is:

\text{H}_2(g) \rightleftharpoons 2\text{H}^+(aq) + 2e^-

Using the Nernst equation:

E = E^{\circ} - \frac{2.303RT}{nF} \log \left( \frac{[\text{H}^+]^2}{P_{\text{H}_2}} \right)

For the standard hydrogen electrode, $E^{\circ} = 0$ , $P_{\text{H}_2} = 1 \, \text{atm}$ , and at $\text{pH} = 3$ ,

[\text{H}^+] = 10^{-3}

So the equation simplifies to:

E = -0.059 \times \text{pH}

Substituting $\text{pH} = 3$ :

E = -0.059 \times 3 \, \text{V}

E = -0.177 \, \text{V}

Hence,

E = -17.7 \times 10^{-2} \, \text{V}

Therefore, $x = 17.7 \approx 18$ , so the answer is $18$ .

Using pH directly

Given: $E^0 = 0$ for the standard hydrogen electrode and $\text{pH} = 3$ , so

[\text{H}^+] = 10^{-3} \, \text{M}

Find: the numerical value of $x$ .

Using the relation stated in the solution:

E = E^0 - 0.059 \times \log(10^3)

E = 0 - 0.059 \times 3

E = -0.177 \, \text{V}

Rewriting in the required form:

-0.177 \, \text{V} = -17.7 \times 10^{-2} \, \text{V}

Thus the required value is $18$ after rounding to the nearest integer.

Common mistakes

Using $[\text{H}^+] = 10^{3}$ instead of $10^{-3}$ . This is wrong because $\text{pH} = -\log[\text{H}^+]$ , so a higher pH means lower hydrogen ion concentration. First convert pH correctly, then substitute.
Ignoring the negative sign in the electrode potential. This is wrong because for a hydrogen electrode at $\text{pH} > 0$ , the potential becomes negative relative to SHE. Keep the sign through the Nernst equation before extracting $x$ .
Reporting $-0.177$ as the final answer instead of the value of $x$ . This is wrong because the question asks for the coefficient in $-x \times 10^{-2} \, \text{V}$ . Rewrite $-0.177 \, \text{V}$ as $-17.7 \times 10^{-2} \, \text{V}$ and then round.

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