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Question
Following reaction takes place in the cell:
`Zn(s) + Ag_2O(s)+H_2O(l) -> Zn^{2+}(aq) + 2Ag (s) + 2OH^- (aq)`
Calculate `Delta_r G^0` of the reaction
[Given ; `E^0_(Zn^{2+}//Zn)` = -0.76V
`E_((Zn^{2+}//Zn)) = 0.76V`
`E_(Ag^4//Ag)^0 = 0.80V, 1F = 96,500 C mol^-1 ]`
Solution
`Zn(s)+Ag_2O(s) +H_2O(e)->Zn^2+(aq)+2Ag(s)+20H^-(aq)`
`Zn(s)+2bare->Zn(s)`, `E_(Zn^{2+}//Zn) = -0.76V (\text{Anode})`
`Ag_2O+2bare -> Ag(s) (\text{cathode}), E_(Ag^+//Ag) = 0.80V`
`E_{cell}^0 = E_(Ag^+//Ag)^0 - E_Zn^{2+}//Zn`
[0.80 -(-0.76)]V=(0.80+0.76)V
= 1.56 V
`Delta_r G^circ =ηFE_{cell}^circ,`
n = number of electron exchanged = 2e-
F= 96500 C mol-1
`E_{cell}^0=1.56V`
`Delta_r G^circ = 2xx96500 C mol^-1 xx 1.56V`
= 301080 V C mol-1
= 3.01 × 105 J mol-1
= 3.01 × 102 KJ mol-1
= -301 KJ mol-1
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