For the reaction H_2 (1 , bar) + 2AgCl_{(s)} | vedclass

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(A) The relationship between standard Gibbs free energy change $(\Delta G^o)$ and standard cell potential $(E^o_{cell})$ is given by the formula: $\De

Vedclass · Accepted Answer

$0.25$

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(A) The relationship between standard Gibbs free energy change $(\Delta G^o)$ and standard cell potential $(E^o_{cell})$ is given by the formula: $\Delta G^o = -nFE^o_{cell}$.Here,the reaction is $H_2 + 2AgCl ightarrow 2Ag + 2H^{+} + 2Cl^{-}$.The number of electrons transferred $(n)$ is $2$.The Faraday constant $(F)$ is approximately $96,500 \, C \, mol^{-1}$.Given $\Delta G^o = -48,250 \, J \, mol^{-1}$.Substituting the values: $-48,250 = -(2) 	imes (96,500) 	imes E^o_{cell}$.$E^o_{cell} = \frac{48,250}{2 	imes 96,500} = \frac{48,250}{193,000} = 0.25 \, V$.

For the reaction $H_2 (1 \, bar) + 2AgCl_{(s)} \rightleftharpoons 2Ag_{(s)} + 2H^{+} (0.1 \, M) + 2Cl^{-} (0.1 \, M)$,$\Delta G^o = -48,250 \, J$ at $25 \, ^oC$. The standard emf of cell in which the given reaction takes place is ................. $V$

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