Question

A button cell used in watches function as following
$Zn(s)+A{g}_{2}O(s)+H_{2}O(l)\rightleftharpoons 2Ag(s)+Z{n}^{2+}(aq)+2OH-(aq)$
If half cell potentials are
$Z{n}_{(aq)}^{2+}+2e-\to Zn(s);E^{\circ }=-0.76V$
$A{g}_2{O}_{(s)}+H_2{O}_{(l)}+2e-\to 2A{g}_{(s)}+2O{H}_{(aq)}^{-};$
$E^{\circ }=0.34V$

• A. 1.34 V
• B. 1.10 V
• C. 0.42 V
• D. 0.84 V

Answer 1

Answer: (B)

$Z{n}^2+(aq)+2e^{-}⟶Zn(s);E^{\circ }=-0\cdot 76V$
$A{g}_{2}O(s)+H_{2}O(l)+2e^{-}⟶2Ag(s)+2OH-(aq)$
$E^{\circ }=0.34V$
$Zn(s)+A{g}_{2}O(s)+H_{2}O(l)⟶2Ag(s)+Z{n}^{+2}(aq)+2OH-(aq);E_{\text{cell }}=?$
$E_{\text{cell }}^{\circ }={\left(E_{R\cdot P}^{\circ }\right)}_{\text{cathode }}-{\left(E_{R\cdot P\cdot }^o\right)}_{\text{anode }}$
$E_{\text{cell }}^{\circ }=0\cdot 34-(-0\cdot 76)=1\cdot 10V$
$E_{\text{cell }}=E_{\text{cell }}^{\circ }=1\cdot 10V$