Consider the following electrochemical cell at standard condition.
Au( s)|QH₂, Q| NH₄ X(0.01 M)| | Ag⁺(1 M) | Ag( s) E_(cell)=+0.4 V
The couple QH₂ / Q represents quinhydrone electrode, the half cell reaction is given below:
[ Given : E_(Ag⁺ / Ag)⁰=+0.8 V and (2.303 RT)/F=0.06 V]
The pK_b value of the ammonium halide salt (NH₄ X) used here is __________ . (nearest integer)

Solution

Chemistry ma’am · 5 steps

Let the cell voltage tell you the pH first; the hydrolysis formula then gives you the pKb.

1. Anode and cathode. The quinhydrone electrode on the left is the anode, so it is oxidised: QH₂ → Q + 2H⁺ + 2e⁻. Silver is reduced at the cathode, so E⁰_(cell) = 0.8 - 0.7 = 0.1 V.
2. Write the Nernst equation. The overall reaction is QH₂ + 2Ag⁺ → Q + 2H⁺ + 2Ag with n = 2. Quinhydrone holds equal amounts of Q and QH₂, so they cancel, and [Ag⁺] = 1 M.

   E_(cell) = 0.1 - 0.06/2 log[H⁺]²/[Ag⁺]² = 0.1 + 0.06 pH

3. Find the pH. 0.4 = 0.1 + 0.06 pH, so pH = 0.3/0.06 = 5.
4. Salt hydrolysis. NH₄X is a salt of a weak base and a strong acid, so pH = 7 - 1/2pK_b - 1/2 log C. With C = 0.01 M, log C = -2, so pH = 8 - 1/2pK_b.
5. So. 5 = 8 - 1/2pK_b gives pK_b = 6.

The common slip: 4. Dropping the -1/2 log C term leaves pH = 7 - 1/2pK_b, which gives a pK_b of 4 instead of 6.

The idea to keep. A quinhydrone electrode is a pH meter in disguise: its potential shifts by 0.06 V per pH unit. For a salt of a weak base and a strong acid, use pH = 7 - 1/2pK_b - 1/2 log C.

Answer from the NTA final answer key. Solution by PrepWiser.