A button cell used in watches functions as following Zns - Ag2Os -H2Ol - 2Ags - Zn2- aq -20H- aq If half cell potentials are Zn2-aq -2e- - Zns E- - -0-76V Ag2Os -H2Ol -2e- -2Ags -2OH-aq E- - 0-34 V the cell potential will be -NEET-UG-2013

Zn^2+_(aq) +2e^ → Zn_(s); E^∘ = 0.76V Ag_2O_(s) +H_2)O_(l) +2e^ → 2Ag_(s) +2OH^ _(aq); E^∘ = 0.34 V Zn_(s) +Ag_2O_(s) +H_2O_(l)→ 2Ag_(s) +Zn^+2_(aq) +2OH^ ...

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Byju's Answer

Standard XII

Chemistry

Equivalent, Molar Conductivity and Cell Constant

A button cell...

Question

A button cell used in watches functions as following $Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ 2 A g (s) + Z n^{2 +} (a q) + 20 H^{-} (a q)$ If half cell potentials are $Z n_{(a q)}^{2 +} + 2 e^{-} \to Z n_{(s)} E^{\circ} = - 0.76 V$ $A g_{2} O_{(s)} + H_{2} O_{(l)} + 2 e^{-} \to 2 A g (s) + 2 O H_{(a q)}^{-} E^{\circ} = 0.34 V$ the cell potential will be :
(  NEET-UG-2013)

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 $1.10 V$
$Z n_{(a q)}^{2 +} + 2 e^{-} \to Z n_{(s)}; E^{\circ} = - 0.76 V$ $A g_{2} O_{(s)} + H_{2}) O_{(l)} + 2 e^{-} \to 2 A g_{(s)} + 2 O H_{(a q)}^{-}; E^{\circ} = 0.34 V$ $Z n_{(s)} + A g_{2} O_{(s)} + H_{2} O_{(l)} \to 2 A g_{(s)} + Z n_{(a q)}^{+ 2} + 2 O H_{(a q)}^{-}; E_{c e l l} = ?$ $E_{c e l l}^{\circ} = (E_{R . P}^{\circ})_{c a t h o d e} - (E_{R . P .}^{\circ})_{a n o d e}$ $E_{c e l l}^{\circ} = 0.34 - (- 0.76) = 1.110 V$ $E_{c e l l}^{\circ} = 1.10 V$

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Similar questions

Q. In the button cells widely used in watches and other devices the following reaction takes place:

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)

Determine

Δ_{r} G^{θ}

and

E^{θ}

for the reaction.

In the button cells widely used in watches and other devices the following reaction takes place:

Zn(s) + Ag₂O(s) + H₂O(l) → Zn²⁺(aq) + 2Ag(s) + 2OH⁻(aq)

Determine and for the reaction.

Consider the cell:

Z n | Z n^{2 +} (a q) (1.0 M) | | C u^{2 +} (a q) (1.0 M) | | C u

The standard reduction potential are

0.350

V for

C u^{2 +} (a q) + 2 e^{-} \to C u

and

- 0.763

V for

Z n^{2 +} (a q) + 2 e^{-} \to Z n

The cell reaction is:

Q. The standard reduction potentials at

298 K

for the following half-cell reactions are given by

{Z n}^{2 +} (a q) + 2 e^{-} ⇌ Z n (s) - 0.762

{C r}^{3 +} (a q) + 3 e^{-} ⇌ C r (s) - 0.74

2 H^{+} (a q) + 2 e^{-} ⇌ H_{2} (g) + 0.00

{F e}^{3 +} (a q) + e^{-} ⇌ {F e}^{2 +} (a q) + 0.77

Which one of the following is the strongest reducing agent?

Q. 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^{\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

The cell potential will be:

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The correct option is A 1.10 VZn2+(aq)+2e−→Zn(s); E∘=−0.76V Ag2O(s)+H2)O(l)+2e−→2Ag(s)+2OH−(aq); E∘=0.34 V Zn(s)+Ag2O(s)+H2O(l)→2Ag(s)+Zn+2(aq)+2OH−(aq);Ecell=? E∘cell=(E∘R.P)cathode−(E∘R.P.)anode E∘cell=0.34−(−0.76)=1.110 V E∘cell=1.10 V