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Question: pH of resulting solution in cell will be ____....

pH of resulting solution in cell will be ____.

Answer

9

Explanation

Solution

To determine the pH of the resulting solution, we first need to identify the reactions occurring at the electrodes during the electrolysis of aqueous KBr using inert platinum electrodes.

The species present in the aqueous KBr solution are K⁺(aq), Br⁻(aq), and H₂O(l).

At the Cathode (Reduction): Possible reduction reactions are:

  1. K⁺(aq) + e⁻ → K(s) ; E∘=−2.92 VE^\circ = -2.92 \, \text{V}
  2. 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) ; E∘=−0.83 VE^\circ = -0.83 \, \text{V} (at pH 7)

Comparing the standard reduction potentials, water has a higher (less negative) reduction potential than K⁺ ions. Therefore, water will be preferentially reduced at the cathode. Cathode Reaction: 2H2O(l)+2e−→H2(g)+2OH−(aq)2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq) This reaction produces hydroxide ions (OH⁻), which will increase the pH of the solution.

At the Anode (Oxidation): Possible oxidation reactions are:

  1. 2Br⁻(aq) → Br₂(l) + 2e⁻ ; Eoxidation∘=−1.09 VE^\circ_{\text{oxidation}} = -1.09 \, \text{V} (or Ereduction∘=+1.09 VE^\circ_{\text{reduction}} = +1.09 \, \text{V})
  2. 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ ; Eoxidation∘=−1.23 VE^\circ_{\text{oxidation}} = -1.23 \, \text{V} (or Ereduction∘=+1.23 VE^\circ_{\text{reduction}} = +1.23 \, \text{V})

Comparing the standard oxidation potentials, Br⁻ ions have a less negative (more positive) oxidation potential than water. Therefore, Br⁻ ions will be preferentially oxidized at the anode. Anode Reaction: 2Br−(aq)→Br2(l)+2e−2Br^-(aq) \rightarrow Br_2(l) + 2e^-

Calculations:

  1. Calculate the total charge passed (Q): Current (I) = 9.65 A Time (t) = 10 s Q=I×t=9.65 A×10 s=96.5 CQ = I \times t = 9.65 \, \text{A} \times 10 \, \text{s} = 96.5 \, \text{C}

  2. Calculate the moles of electrons transferred (nen_e): Using Faraday's constant (F = 96500 C/mol e⁻): ne=QF=96.5 C96500 C/mol e−=0.001 mol e−n_e = \frac{Q}{F} = \frac{96.5 \, \text{C}}{96500 \, \text{C/mol e}^-} = 0.001 \, \text{mol e}^-

  3. Calculate the moles of OH⁻ produced: From the cathode reaction, 2H2O(l)+2e−→H2(g)+2OH−(aq)2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq), we see that 2 moles of electrons produce 2 moles of OH⁻ ions. Therefore, moles of OH⁻ produced = moles of electrons transferred = 0.001 mol0.001 \, \text{mol}.

  4. Calculate the concentration of OH⁻ ions ([OH⁻]): Volume of the solution = 100 L [OH−]=moles of OH−Volume of solution=0.001 mol100 L=1×10−5 M[OH^-] = \frac{\text{moles of OH}^-}{\text{Volume of solution}} = \frac{0.001 \, \text{mol}}{100 \, \text{L}} = 1 \times 10^{-5} \, \text{M}

  5. Calculate pOH: pOH=−log⁡[OH−]=−log⁡(1×10−5)=5pOH = -\log[OH^-] = -\log(1 \times 10^{-5}) = 5

  6. Calculate pH: At 25°C, pH+pOH=14pH + pOH = 14 pH=14−pOH=14−5=9pH = 14 - pOH = 14 - 5 = 9

The pH of the resulting solution will be 9.

Explanation of the solution:

  1. Identify cathode reaction: 2H2O(l)+2e−→H2(g)+2OH−(aq)2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq) (water is reduced preferentially over K⁺).
  2. Calculate charge passed: Q=I×t=9.65 A×10 s=96.5 CQ = I \times t = 9.65 \, \text{A} \times 10 \, \text{s} = 96.5 \, \text{C}.
  3. Calculate moles of electrons: ne=Q/F=96.5 C/96500 C/mol=0.001 moln_e = Q/F = 96.5 \, \text{C} / 96500 \, \text{C/mol} = 0.001 \, \text{mol}.
  4. Determine moles of OH⁻ produced: From cathode reaction stoichiometry, moles of OH⁻ = moles of electrons = 0.001 mol0.001 \, \text{mol}.
  5. Calculate [OH⁻]: [OH−]=0.001 mol/100 L=1×10−5 M[OH^-] = 0.001 \, \text{mol} / 100 \, \text{L} = 1 \times 10^{-5} \, \text{M}.
  6. Calculate pOH: pOH=−log⁡(1×10−5)=5pOH = -\log(1 \times 10^{-5}) = 5.
  7. Calculate pH: pH=14−pOH=14−5=9pH = 14 - pOH = 14 - 5 = 9.