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Calculate the minimum weight of NaOH required to be added in R.H.S to consume all the $H^+$ present in R.H.S of cell of emf to .701V at $25^{\large\circ}C$.Given: $Zn/Zn^{+2}(0.1M)\parallel HCl$(1litre)$/H_2/$(1atm)/P:-$E^0_{Zn/Zn^{+2}}=0.7602$

$\begin{array}{1 1}(a)\;2.3gm\\(b)\;1.92gm\\(c)\;1.26gm\\(d)\;3.26gm\end{array}$

1 Answer

For calculating the minimum weight of NaOH ,which is supposed to be added to cathode compartment ,we are required to know the $[H^+]$ present in this compartment.
There occuring in the cell is
$Zn+2H^+\rightarrow Zn^{+2}+H_2(g)$
$E_{cell}=E^0_{H^+/H_2}-E^0_{Zn^{+2}/Zn}-\large\frac{0.059}{2}$$\log \large\frac{P_{H_2}[Zn^+2]}{[H^+]^2}$
$0.701=0.76-\large\frac{0.059}{2}$$\log \large\frac{1\times 0.1}{[H^+]^2}$
$\therefore [H^+]=0.0316$mol/lit
Since sufficient NaOH is to be added to cathode compartment to consume entire $H^+$,the equivalent of HCl must be equal to eq of NaOH.Let the weight of NaOH added be $x$gm.
So $x=1.264$gm
Hence (c) is the correct answer.
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