Question

The emf of the cell, $Pt|H_2\left(1\text{atm}\right),|H^{+}(0.1\text{M},40\text{mL})||A{g}^{+}\left(0.8\text{mM}\right)|Ag$ is $0.9\text{V}$. Calculate the emf when $60\text{mL}$ of $0.05\text{M}NaOH$ is added to the anodic compartment.
Given: $\frac{2.303RT}{F}=0.06,\log 2=0.3$

Answer 1

The cell given here is,
$Pt|H_2\left(1\text{atm}\right),|H^{+}(0.1\text{m},40\text{mL})||A{g}^{+}\left(0.8\text{M}\right)|Ag$
So according to Nernst equation,
$E_{cell}=E_{cell}^{\circ }-\frac{2.303RT}{nF}\log \frac{\left[H^{+}\right]}{\left[A{g}^{+}\right]}$
$\Rightarrow 0.9=E_{cell}^{\circ }-\frac{0.06}{1}log\frac{0.1}{0.8}$
$\Rightarrow 0.9=E_{cell}^{\circ }+0.054$
$\Rightarrow E_{cell}^{\circ }=0.846\text{V}$
The molarity of $H^{+}$ ions after addition of $NaOH$
$=\frac{M_1{V}_1-M_2{V}_2}{V_1+V_2}=\frac{0.1\times 40-0.05\times 60}{40+60}$
=$\frac{(4-3)\times {10}^{-3}}{100}=0.01\times {10}^{-3}\text{M}$
Now, EMF of cell-
$E_{cell}=-E_{\text{cell}}^{\circ }-\frac{0.06}{1}\log \frac{\left(0.01\right)}{\left(0.8\right)}$
$\Rightarrow E_{\text{cell}}=0.846-0.06\log \frac{\left(0.01\right)}{\left(0.8\right)}$
$\Rightarrow E_{\text{cell}}=0.846+0.06\log \frac{\left(0.8\right)}{\left(0.01\right)}$
$\Rightarrow E_{\text{cell}}=0.846+0.06\log 80$
$\Rightarrow E_{\text{cell}}=0.846+0.06\times 1.90$
$\Rightarrow E_{\text{cell}}=0.96\text{V}$