QH2+2Ag+→2Ag+Q+2H+E=E∘−20.06log[H+]2E=E∘−0.06×log[H+]pH=−log(H+)=0.06E−Eo=0.060.4−0.1=0.060.3=5
pH+NH4X=7−21pKb−21logC5=7−21×pKb−21log(10−2)pKb=6
JEE Main 2025 — Chemistry Physical Chemistry
Consider the following electrochemical cell at standard condition.
Au(s)∣QH2,Q∣NH4X(0.01M)∣∣Ag+(1M)∣Ag(s)Ecell =+0.4 V
The couple QH2/Q represents quinhydrone electrode, the half cell reaction is given below

The pKb value of the ammonium halide salt (NH4X) used here is _________. (nearest integer)
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