MCQ
The standard electrode potential for $Cu^{+2}/Cu$ is $0.34\, volt$ calculate the reduction potential at $pH = 14$ for the above couple ............... $\mathrm{V}$ $[K_{sp}[Cu(OH)_2) = 1 \times 10^{-19}]$
- A$+0.03$
- B$-3.09$
- ✓$-0.22$
- D$1.50$
$E=E^{\circ}-\frac{0.06}{2} \log \frac{1}{\left(\mathbf{C u}^{+2}\right)}$
$\mathrm{Cu}(\mathrm{OH})_{2} \rightleftharpoons \mathrm{Cu}^{+2}+2 \mathrm{OH}^{-}$
$\mathrm{k}_{\mathrm{sp}}=\left(\mathrm{Cu}^{+2}\right)\left(\mathrm{OH}^{-}\right)^{2}$
$10^{-19}=\left(\mathrm{Cu}^{+2}\right)(1)^{2}$
$\left( {{\text{C}}{{\text{u}}^{ + 2}}} \right) = {10^{ - 19}}$ ${{\text{P}}^{\text{H}}} = 14$
${\mathrm{P}^{\mathrm{OH}}=0}$
${\mathrm{OH}^{-} \Rightarrow 1}$
solve it
${E}=-\,0.22\, \mathrm{V}$
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($A$) a high activation energy usually implies a fast reaction.
($B$) rate constant increases with increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy.
($C$) higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant.
($D$) the pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy.