The $$zinc/silver$$ oxide cell is used in electric watches. The reaction is as following,$$Zn2++2e-$$→Zn;E°$$=-0.760$$ $$VAg2O+H2O+2e-$$→$$2Ag+2OH-$$;E°$$=0.344$$ V If F is $$96$$,$$500Cmol-1$$,ΔG° of the cell will be

Question

The $$zinc/silver$$ oxide cell is used in electric watches. The reaction is as following,$$Zn2++2e-$$→Zn;E°$$=-0.760$$ $$VAg2O+H2O+2e-$$→$$2Ag+2OH-$$;E°$$=0.344$$ V If F is $$96$$,$$500Cmol-1$$,ΔG° of the cell will be

Infinity Learn · Accepted Answer

Δ$$G0$$ $$=$$ −$$nFE0From$$ the reaction, n $$=$$ $$2$$Δ$$G0$$ $$=$$ −$$296500$$   Ecello  .........(1)Ecello$$ $$=$$   Ecathodeo  −     $$EanodeoZn+2$$  $$+$$  $$2e$$−  →   Zn,    $$E0=$$−$$0.76VAg2O+H2O+2e$$¯  →    $$2Ag+2OH$$−, $$E0=+0.344VSRP$$ value of $$Ag2O$$  half cell is high ∴ it acts like cathodeThus …   Anode ; Zn→$$Zn+2$$ $$+2e-$$;  Cathode; $$Ag2O$$ $$+H2O$$ $$+2e-$$→ $$2Ag$$ $$+2OH-Zn$$ $$+Ag2O+H2O$$ →   $$Zn+2$$    $$+$$   $$2Ag$$  $$+$$  $$2OH$$− Ecello $$=$$   $$+0.344$$−−$$0.76=$$ $$+$$ $$1.104$$ VSubstituting  the value in $$---(1)Δ$$G0$$ $$=$$ −$$296.5$$   $$1.104kJ=$$ –$$213.072$$ kJ

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