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Question: \(\Delta Gº\) for the cell with the cell reaction: \[Zn_{(s)} + Ag_{2}O_{(s)} + H_{2}O_{(l)} \right...

ΔGº\Delta Gº for the cell with the cell reaction:

Zn(s)+Ag2O(s)+H2O(l)→Zn2+(aq)+2Ag(s)+2OH−(aq)[Eºag2O/Ag=0.344V,EºZn2+/Zn=−0.76V]Zn_{(s)} + Ag_{2}O_{(s)} + H_{2}O_{(l)} \rightarrow Z{n^{2 +}}_{(aq)} + 2Ag_{(s)} + 2O{H^{-}}_{(aq)}\lbrack Eº_{ag_{2}O/Ag} = 0.344V,Eº_{Zn^{2 +}/Zn} = - 0.76V\rbrack

A

2.13×105Jmol−12.13 \times 10^{5}Jmol^{- 1}

B

−2.13×105Jmol−1- 2.13 \times 10^{5}Jmol^{- 1}

C

1.06×105Jmol−11.06 \times 10^{5}Jmol^{- 1}

D

−1.06×105Jmol−1- 1.06 \times 10^{5}Jmol^{- 1}

Answer

−2.13×105Jmol−1- 2.13 \times 10^{5}Jmol^{- 1}

Explanation

Solution

EºCell=EºAg2O/Ag−EºZn2+/ZnEº_{Cell} = Eº_{Ag_{2}O/Ag} - Eº_{Zn^{2 +}/Zn}

=0.344−(−0.76)=1.104V= 0.344 - ( - 0.76) = 1.104V

ΔGº=−nFEºcell=−2×96500×1.104=−2.13×105Jmol−1\Delta Gº = - nFEº_{cell} = - 2 \times 96500 \times 1.104 = - 2.13 \times 10^{5}Jmol^{- 1}