એક્યુટી 3 ઇલેક્ટ્રોકીમિસ્ટ્રી (ઇનટેક્સ્ટ પ્રશ્નો-2)

જવાબ

હાઇડ્રજન ઇલેક્ટ્રોડ માટે, ${H}^{+}+{e}^{-} \longrightarrow \frac{1}{2} {H_2} \text {, it is given that } {pH}=10$

$\therefore\left[{H}^{+}\right]=10^{-10} {M}$

હવે, નર્ન્સ્ટ સમીકરણનો ઉપયોગ કરીને:

$ {H_{({H}^{+} / \frac{1}{2} {H_2} )}}=E_{({H}^{+} \ \frac{1}{2} {H_2} )}^{\ominus}-\frac{{R} T}{n {~F}} \ln \frac{1}{ [{H}^{+} ]} $

$\quad\quad\quad\quad\quad\quad =E_{({H}^{+} / \frac{1}{2} {H_2})}^{\ominus}-\frac{0.0591}{1} \log \frac{1}{[{H}^{+}]} $

$\quad\quad\quad\quad\quad\quad=0-\frac{0.0591}{1} \log \frac{1}{[10^{-10}]} $

$\quad\quad\quad\quad\quad\quad=-0.0591 \times 10 $

$\quad\quad\quad\quad\quad\quad=-0.591 {~V}$

જવાબ

નર્ન્સ્ટ સમીકરણનો ઉપયોગ કરીને પરિણામ મળ્યું છે:

$$ \begin{aligned} & E_{\text {(cell) }}=E_{\text {(cell) }}^{\ominus}-\frac{0.0591}{n} \log \frac{\left[{Ni}^{2+}\right]}{\left[{Ag}^{+}\right]^{2}} \\ \\ & \quad\quad\quad\quad=1.05-\frac{0.0591}{2} \log \frac{(0.160)}{(0.002)^{2}} \\ \\ & \quad\quad\quad\quad=1.05-0.02955 \log \frac{0.16}{0.000004} \\ \\ & \quad\quad\quad\quad=1.05-0.02955 \log 4 \times 10^{4} \\ \\ & \quad\quad\quad\quad=1.05-0.02955(\log 10000+\log 4) \\ \\ & \quad\quad\quad\quad=1.05-0.02955(4+0.6021) \\ \\ & \quad\quad\quad\quad=0.914 {~V} \end{aligned} $$

જવાબ

અહીં, $n=2, E_{\text {cell }}^{\ominus}=0.236 {~V},{ _{T}}=298 {~K}$

અમે જાણીએ છીએ કે:

$\Delta_{r} {G}^{\ominus}=-n {FE_\text {cell }}^{\ominus}$

$\quad\quad\quad\quad=-2 \times 96487 \times 0.236$

$\quad\quad\quad\quad=-45541.864 {~J} {~mol}^{-1}$

$\quad\quad\quad\quad=-45.54 {~kJ} {~mol}^{-1}$

ફરીથી, $\Delta_r G^{\ominus}= -2.303 R T \log K_{c}$

$\quad\Rightarrow \log K_{{c}}=-\frac{\Delta_{r} G^{\ominus}}{2.303 {R} T}$

$\quad\quad\quad\quad\quad\quad =-\frac{-45.54 \times 10^{3}}{2.303 \times 8.314 \times 298} $

$\quad\quad\quad\quad\quad\quad=7.981$

$\therefore K_{{c}}=$ એન્ટિલોગ (7.981)

$\quad\quad\quad\quad=9.57 \times 10^{7}$