A button cell used in watches functions as following:

$Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ 2 A g (s) + Z n^{2 +} (a q) + 2 O H^{-} (a q)$

If half cell potentials are: $Z n^{2 +} (a q) + 2 e \to Z n (s); E^{0} = - 0.76 V$

$A g_{2} O (s) + H_{2} O (l) 2 e \to 2 A g (s) + 2 O H^{-} (a q), E^{0} = 0.34 V$

The cell potential will be:

1.10 V

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0.42 V

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0.84 V

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1.34 V

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Solution

The correct option is A $1.10 V$
Anode is always the site of oxidation thus anode half cell reaction is-
$Z n^{2 +} + 2 e^{-} \to Z n (s); E^{o} = - 0.76 V$
Cathode half cell reaction is-
$A g_{2} O + H_{2} O + 2 e^{-} \to 2 A g (s) + 2 O H^{-} (a q); E^{o} = 0.34 V$

$E o_{c e l l} = E_{c a t h o d e}^{o} - E_{a n o d e}^{o} = 0.34 - (- 0.76) = 1.10 V$

Hence, the correct option is $A$

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Similar questions

Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ 2 A g (s) + Z n^{2 +} (a q) + 2 O H^{-} (a q)

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{\circ} = - 0.76 V

A g_{2} O (s) + H_{2} O (l) + 2 e^{-} \to 2 A g (s) + 2 O H^{-} (a q), E^{\circ} = 0.34 V

△ G^{\circ}

Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ Z n^{2 +} (a q) + 2 A g (s) + 2 O H^{-} (a q)

E_{A g^{+} / A g}^{\circ} = + 0.80 V

E_{Z n^{2 +} / Z N}^{\circ} = - 0.76 V

Δ_{r} G^{o}

Z n_{(s)} + A g_{2} O_{(s)} + H_{2} O_{(l)} \to Z n_{(a q)}^{2 +} + 2 A g_{(s)} + 2 O H_{(a q)}^{-}

[E_{A g_{2} O / A g}^{o} = 0.344 V, E_{Z n^{2 +} / Z n}^{o} = - 0.76 V]

Δ G

Z n_{(s)} + A g_{2} O_{(s)} + H_{2} O_{(l)} \to Z n_{(a q)}^{2 +} + 2 A g_{(s)} + 2 O H^{-}_{(a q)}

E_{{A g}^{+} / A g}^{0} = + 0.80 V

E_{{Z n}^{+ 2} / Z n}^{0} = - 0.76 V

Z n (s) + A g_{2} O (s) + H_{2} O (l) \to Z n^{2 +} (a q) + 2 A g (s) + 2 O H^{-} (a q)

Δ_{r} G^{θ}

E^{θ}

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