Question 4 of 29intermediate🔧 ApplyNumerical3 marks

Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Correct Answer

Potential of hydrogen electrode = -0.591 V

The hydrogen electrode reaction is: H⁺(aq) + e⁻ → ½H₂(g)

For hydrogen electrode, the Nernst equation at 298 K gives: E = E⁰ + 0.0591 log[H⁺]

Since E⁰ for standard hydrogen electrode = 0 V

Given: pH = 10 Therefore, log[H⁺] = -pH = -10

Substituting: E = 0 + 0.0591 × (-10) E = -0.591 V

Hence, the potential of hydrogen electrode in contact with a solution of pH 10 is -0.591 V.

Exercise: Intext Questions | Q: 2.4 | (Page Page 11)
For More Understanding

Explanation

This question tests the application of Nernst equation to hydrogen electrode. The context provides the half-cell reaction for hydrogen electrode (H⁺ + e⁻ → ½H₂) and establishes that standard hydrogen electrode potential E⁰ = 0 V. The Nernst equation relates electrode potential to ion concentration. At 298 K, for hydrogen electrode: E = E⁰ + 0.0591 log[H⁺]. Since pH = -log[H⁺], when pH = 10, log[H⁺] = -10. Substituting gives E = -0.591 V, indicating the electrode is negative relative to standard hydrogen electrode due to very low H⁺ concentration.

Solution Steps

  1. Step 1: Write the hydrogen electrode half-reaction: H⁺(aq) + e⁻ → ½H₂(g)

  2. Step 2: Apply Nernst equation at 298 K: E = E⁰ + 0.0591 log[H⁺]

  3. Step 3: Since E⁰ for standard hydrogen electrode = 0 V, the equation becomes: E = 0.0591 log[H⁺]

  4. Step 4: Calculate log[H⁺] from pH: pH = 10, so log[H⁺] = -10

  5. Step 5: Substitute and calculate: E = 0.0591 × (-10) = -0.591 V

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