$E_{cell}^o = E_{{\rm{cathode}}}^o - E_{{\rm{anode}}}^o$ $= -0.44 -(-0.14)= -0.30\,V$
The negative $EMF$ suggests that the reaction goes spontaneously in reversed direction.
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Product in above reaction is
$\mathrm{A}+\mathrm{B} \underset{\text { Step } 3}{\text { Step } 1} \mathrm{C} \xrightarrow{\text { Step } 2} \mathrm{P}$
Some details of the above reaction are listed below.
| Step |
Rate constant $\left(\sec ^{-1}\right)$ |
Activation energy $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$ |
| $1$ | ${k}_1$ | $300$ |
| $2$ | ${k}_2$ | $200$ |
| $3$ | ${k}_3$ | $\mathrm{Ea}_3$ |
If the overall rate constant of the above transformation (k) is given as $\mathrm{k}=\frac{\mathrm{k}_1 \mathrm{k}_2}{\mathrm{k}_3}$ and the overall activation energy $\left(E_2\right)$ is $400 \mathrm{~kJ} \mathrm{~mol}^{-1}$, then the value of $\mathrm{Ea}_3$ is $\qquad$ $\mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer)
[Assume $100 \%$ ionisation of the complex and $CaCl _{2},$ coordination number of $Cr$ as $6,$ and that all $NH _{3}$ molecules are present inside the coordination sphere $]$