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Byju's Answer

Standard XII

Chemistry

Nernst Equation

Calculate the...

Question

Calculate the EMF of the cell at 298 K.
$P t | H_{2} (1 a t m) | N a O H (x M) | | N a C l (x M) | A g C l (s) | A g$
If $E_{C l^{-} / A g C l / A g}^{0} = 0.222 V$ .

1.048 V

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-0.04 V

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-0.604 V

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emf depends on x and cannot be determined unless value of x is given

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Solution

The correct option is A 1.048 V
Anode: $\frac{1}{2} H_{2} \to H^{+} + 1 e^{-}$
Cathode: $A g C l + 1 e^{-} \to A g + C l^{-}$
-----------------------------------------------------------
net: $\frac{1}{2} H_{2} + A g C l 1 e^{-} - - \to H^{+} + A g + C l^{-}$
$E_{c e l l} = + 0.222 + \frac{0.059}{1} l o g \frac{1}{[H^{+}] [C l^{-}]}$
$= + 0.222 + 0.059 l o g \frac{[O H^{-}]}{[10^{- 14}] [C l^{-}]} = + 0.222 + 0.059 (14) = + 1.048 volt$

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Calculate the EMF of the cell at 298 K.Pt|H2(1atm)|NaOH(xM)||NaCl(xM)|AgCl(s)|AgIf E0Cl−/AgCl/Ag=0.222V.

The correct option is A 1.048 VAnode: 12H2→H++1e−Cathode: AgCl+1e−→Ag+Cl−-----------------------------------------------------------net: 12H2+AgCl1e−−−→H++Ag+Cl−Ecell=+0.222+0.0591log1[H+][Cl−]=+0.222+0.059log[OH−][10−14][Cl−]=+0.222+0.059(14)=+1.048 volt

Calculate the EMF of the cell at 298 K.
$P t | H_{2} (1 a t m) | N a O H (x M) | | N a C l (x M) | A g C l (s) | A g$
If $E_{C l^{-} / A g C l / A g}^{0} = 0.222 V$ .