পাঠ ৩ ইলেক্ট্ৰ'কেমিষ্ট্ৰি (অন্তঃপাঠ্য প্ৰশ্ন-২)

উত্তৰ

হাইড্ৰজেন ইলেক্ট্ৰ’ডৰ বাবে, ${H}^{+}+{e}^{-} \longrightarrow \frac{1}{2} {H_2} \text {, it is given that } {pH}=10$

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

এতিয়া, নাৰ্নষ্ট সমীকৰণ ব্যৱহাৰ কৰি:

$ {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 $

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

উত্তৰ

নাৰ্নষ্ট সমীকৰণ প্ৰয়োগ কৰি আমি পাইছো:

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

উত্তৰ

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

আমি জানো যে:

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

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

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

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

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

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

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

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

$\therefore K_{{c}}=$ এণ্টিলগ (৭.৯৮১)

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