Question

Equinormal solution of two weak acids, $HA(p{K}_a=3)$ and $HB(p{K}_a=5)$ are each placed in contact with standard hydrogen electrode at ${25}^{\circ }C$. When a cell is constructed by interconnecting them through a salt bridge, find the e.m.f of the cell.

• A. $E=0.118V$
• B. $E=0.059V$
• C. $E=-0.059V$
• D. None of these

Answer 1

The correct option is B $E=0.059V$

The cell is,
$Pt{H}_{2_{1atm}}\left|H{A}_2\right|\left|H{A}_1\right|H_{2_{1atm}}Pt$
At L.H.S. : $E_{H/H^{+}}=E_{O{P}_{H/H^{+}}}^{\circ }+\frac{0.059}{1}{\log }_{10}[H^{+}{]}_2$
$\because -\log H^{+}=pH$ $\therefore E_{H/H^{+}}=E_{O{P}_{H/H^{+}}}^{\circ }-0.059(pH{)}_2$
At R.H.S. : $E_{H^{+}/H}=E_{R{P}_{H^{+}/H}}^{\circ }+\frac{0.059}{1}{\log }_{10}[H^{+}{]}_1$
$\therefore E_{H^{+}/H}=E_{R{P}_{H^{+}/H}}^{\circ }-0.059(pH{)}_1$
For Acid $H{A}_1$ $H{A}_1\rightleftharpoons H^{+}+A_1^{-}$
$\left[H^{+}\right]=C.\alpha =\sqrt{K_a.C}$
$\therefore (pH{)}_1=\frac{1}{2}p{K}_{a_1}-\frac{1}{2}{\log }_{10}C$
$(pH{)}_2=\frac{1}{2}p{K}_{a_2}-\frac{1}{2}{\log }_{10}C$
($\because$ C are same )
$\therefore E_{cell}=E_{O{P}_{H^{+}/H}}+E_{P{R}_{H^{+}/H}}$
for II for I
$=0.059[\frac{1}{2}p{K}_{a_2}-\frac{1}{2}p{K}_{a_1}]=\frac{0.059}{2}[5-3]$
$=+0.059$