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Chemistry Question on Electrochemistry

A button cell used in watches function as following Zn(s) + Ag2O(s) + H2O(l) ⇌\rightleftharpoons 2Ag(s) + Zn2+(aq) + 2OH-(aq). If half cell potentials are Zn(aq)2++2e→Zn(s);E∘=−0.76V{Zn^{2+}_{ (aq)} + 2e\to Zn(s); E^{\circ} = -0.76V} Ag2O(s)+H2O(l)+2e→2Ag(s)+2OH(aq)−;{Ag2O_{(s)} + H2O_{(l)} + 2e\to 2Ag_{(s)} + 2OH^{-}_{(aq)};} E∘=0.34VE^{\circ} = 0.34V

A

1.34 V

B

1.10 V

C

0.42 V

D

0.84 V

Answer

1.10 V

Explanation

Solution

Zn2+(aq)+2e−→Zn(s);E∘=−0.76V{ Zn^{2+}(aq) + 2e^-\to Zn(s); E^{\circ} = -0.76 V}
Ag2O(s)+H2O(l)+2e−→2Ag(s)+2OH−(aq);{Ag2O(s) + H2O(l) + 2e^- \to 2Ag(s) + 2OH^-(aq);}
E∘=0.34 VE^{\circ} = 0.34 \, V
Zn(s)+Ag2O(s)+H2O(l)→2Ag(s)+Zn+2(aq)+2OH−(aq);Ecell=?{Zn(s) + Ag2O(s) + H2O(l) \to 2Ag(s) + Zn^{+2} (aq) + 2OH^-(aq); E_{cell} = ?}
Ecell∘=(ER.P.∘)cathode−(ER.P.∘)anode{ E^{\circ}_{cell} = (E^{\circ}_{R.P.})_{cathode} - (E^{\circ}_{R.P.})_{anode}}
Ecell∘=0.34−(−0.76)=1.10V{E^{\circ}_{cell} = 0.34 - (-0.76) = 1.10 V}
Ecell=Ecell∘=1.10V{E_{cell} = E^{\circ}_{cell} = 1.10V}

So, the correct answer is (B): 1.10 V