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1.26gm 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:-$E^0_{Zn/Zn^{+2}}=0.7602$.The emf of cell after addition of NaOH will be

$\begin{array}{1 1}(a)\;0.3V\\(b)\;0.44V\\(c)\;0.37V\\(d)\;0.52V\end{array}$

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After the addition of sufficient NaOH,the solution become neutral & we have $[H^+]=10^{-7}$
$\therefore E_{cell}=0.76-\large\frac{0.059}{2}$$\log \large\frac{0.1}{(10^{-7})^2}$
$\Rightarrow 0.376$V
Hence (c) is the correct answer.
answered Feb 18, 2014 by sreemathi.v

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