Question

The potential of the cell at ${25}^{\circ }C$ is :
Given $p{K}_b$ of $N{H}_{4}OH=4.74$

$Pt\left|\begin{array}{c}{H}_2\\ \left(1atm\right)\end{array}\right|\begin{array}{c}H{H}_{4}OH\\ \left({10}^{3}M\right)\end{array}|\left|\begin{array}{c}NaOH\\ \left({10}^{-3}M\right)\end{array}\right|\begin{array}{c}{H}_2\\ \left(1atm\right)\end{array}|Pt$

• A. $0.05V$
• B. $0.04V$
• C. $-0.189V$
• D. $0.189V$

Answer 1

The correct option is A $0.05V$

It is concentration cell, therefore, $E^{\ominus }cell=0$
For weak base $N{H}_{4}OH$,
$pOH=\frac{1}{2}(4.74-log{10}^{-3})=3.87$
$\therefore$ pH = 14 3.87 = 10.13
For strong base NaOH,
pOH = 3, pH = 14 - 3 = 11
$\therefore E_{cell}=-0.059(p{H}_c-p{H}_a)$
$=-0.059(11-10.13)$
$=-0.059\times 0.87$
$=-0.05V$