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0 votes
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in Electrochemistry by (430 points)
edited by

\( \Delta G ^{\circ} \) of the cell reaction \( AgCl ( s )+\frac{1}{2} H _{2}( g ) \rightarrow Ag ( s )+ H ^{*}+ Cl ^{-} \)is \( \left.-21.52 k \right) . \Delta G ^{\circ} \) of \( 2 AgCl ( s )+ H _{2}( g ) \rightarrow 2 Ag ( s )+2 H ^{+}+2 Cl ^{-} \)is 

(1) \( -21.52 kJ \) 

(2) \( -10.76 kJ \) 

(3) \( -43.04 kJ \) 

(4) \( 43.04 kJ \)

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1 Answer

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by (44.1k points)

Correct option is (C) 43.04 KJ

\(\overset{+1}{Ag}Cl\) + \(\frac12\)H2(g) \(\longrightarrow\) \(\overset{o}{Ag}(s)\) + H+ + Cl-

\(\Delta \overset{o}G\) = -21.52 KJ

As we know, \(\Delta \overset{o}G\) is  extensive property. It depends on the number of electrons transfer

\(\Delta \overset{o}G\) = -nFEo

(Eo = intensive property) (F = Constant 96500 c)

therefore,

for 

2Ag Cl(s) + H2(g)\(\longrightarrow\) 2Ag(s) + 2H+ + 2Cl-

in this reaction, 2 moles of electrons are transfer

\(\therefore\) \(\Delta \overset{o}G\)  = 2 x -21.52 KJ

 = -43.04 KJ

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