Question
Explain Equilibrium constant for Daniell by nernst equation and calculate it's value
✓
Answer
→ The cell reaction of Daniell cell is represented as : $ Zn ( s )+ Cu ^{2+}( aq ) \rightarrow Zn ^{2+}( aq )+ Cu ( s ) $
→ As time passes, the concentration of $Zn ^{2+}$ keeps on increasing while the concentration of $Cu ^{+2}$ keeps on decreasing.
→ At the same time voltage measured on the voltmeter decreases
→ After some time, it will be observed that there is no change in the concentration of $Cu ^{+2}$ and $Zn ^{+2}$ ions and at the same time, voltmeter shows zero reading.
This indicates that equilibrium has been attained.
$
E _{\text {cell }}= E _{\text {cell }}^0-\frac{0.059}{ n } \log \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}
$
→ At Equilibrium state
$
E _{\text {cell }}=0.0 V, \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}= K _{ c }
$
$
\begin{array}{l}
\therefore 0.0= E _{\text {cell }}^0-\frac{0.059}{ n } \log K _{ C } \\
\therefore E _{\text {cell }}^0=\frac{0.059}{ n } \log K _{ C }
\end{array}
$
For Daniell cell $n =2$
$
\begin{array}{l}
E _{\text {cell }}^0=1.1 V \\
E _{\text {cell }}^0=\frac{0.059}{2} \log K _{ C } \\
\therefore 1.1=\frac{0.059}{2} \log K _{ C } \\
\therefore \frac{1.1 \times 2}{0.059}=\log K _{ C } \\
\therefore \log K _{ C }=37.288 \\
\therefore K _{ C }=2 \times 10^{37}
\end{array}
$
→ As time passes, the concentration of $Zn ^{2+}$ keeps on increasing while the concentration of $Cu ^{+2}$ keeps on decreasing.
→ At the same time voltage measured on the voltmeter decreases
→ After some time, it will be observed that there is no change in the concentration of $Cu ^{+2}$ and $Zn ^{+2}$ ions and at the same time, voltmeter shows zero reading.
This indicates that equilibrium has been attained.
$
E _{\text {cell }}= E _{\text {cell }}^0-\frac{0.059}{ n } \log \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}
$
→ At Equilibrium state
$
E _{\text {cell }}=0.0 V, \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}= K _{ c }
$
$
\begin{array}{l}
\therefore 0.0= E _{\text {cell }}^0-\frac{0.059}{ n } \log K _{ C } \\
\therefore E _{\text {cell }}^0=\frac{0.059}{ n } \log K _{ C }
\end{array}
$
For Daniell cell $n =2$
$
\begin{array}{l}
E _{\text {cell }}^0=1.1 V \\
E _{\text {cell }}^0=\frac{0.059}{2} \log K _{ C } \\
\therefore 1.1=\frac{0.059}{2} \log K _{ C } \\
\therefore \frac{1.1 \times 2}{0.059}=\log K _{ C } \\
\therefore \log K _{ C }=37.288 \\
\therefore K _{ C }=2 \times 10^{37}
\end{array}
$
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