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Question: A cell diagram shown below contains one litre of buffer solution of HA($pK_a$ = 4) and NaA in both c...

A cell diagram shown below contains one litre of buffer solution of HA(pKapK_a = 4) and NaA in both compartments. What is the cell e.m.f.?

A

0.03 V

B

0.06 V

C

-0.06 V

D

None of these

Answer

0.06 V

Explanation

Solution

  1. Determine pH in each half-cell using the Henderson–Hasselbalch equation:

    pH=pKa+log⁡[A−][HA]\text{pH} = pK_a + \log\frac{[\text{A}^-]}{[\text{HA}]}

    • Left half-cell:

      pH=4+log⁡10.1=4+1=5.\text{pH} = 4 + \log\frac{1}{0.1} = 4 + 1 = 5.

    • Right half-cell:

      pH=4+log⁡11=4.\text{pH} = 4 + \log\frac{1}{1} = 4.

  2. Determine electrode potentials using the Nernst equation for the hydrogen electrode:

    For a hydrogen electrode, the half-cell potential is given by:

    E=−0.0591×pH(in volts at 298 K)E = -0.0591 \times \text{pH} \quad (\text{in volts at 298 K})

    • Left electrode potential:

      Eleft=−0.0591×5=−0.2955 VE_{\text{left}} = -0.0591 \times 5 = -0.2955\ \text{V}

    • Right electrode potential:

      Eright=−0.0591×4=−0.2364 VE_{\text{right}} = -0.0591 \times 4 = -0.2364\ \text{V}

  3. Calculate the cell electromotive force (e.m.f.):

    Ecell=Eright−Eleft=(−0.2364 V)−(−0.2955 V)=0.0591 V≈0.06 V.E_{\text{cell}} = E_{\text{right}} - E_{\text{left}} = (-0.2364\ \text{V}) - (-0.2955\ \text{V}) = 0.0591\ \text{V} \approx 0.06\ \text{V}.

Since the electrons flow from the left (more negative) electrode to the right electrode, the overall cell e.m.f. is approximately 0.06 V.