First slide

Standard EMF of a cell

Question

A button cell used in watches, functions as following $\begin{aligned} Zn (s) + {Ag}_{2} O (s) + H_{2} O (l) ⇌ \\ 2 Ag (s) + {Zn}^{2 +} (,) + 2 {OH}^{-} (aq) \end{aligned}$
If half-cell potentials are $\begin{matrix} {Zn}^{2 +} (aq) + 2 e^{-} ⟶ Zn (s); E^{\circ} = - 0 .76 V \\ {Ag}_{2} O (s) + H_{2} O (l) + 2 e^{-} ⟶ 2 Ag (s) + 2 {OH}^{-} \end{matrix}$
$E^{\circ} = 0 .34 V$ the cell potential will be

Moderate

A

1.1 V
B

0.42 V
C

0.84 V
D

1.34 V

Solution

Anode is always the site of oxidation and cathode is the site of reduction, Thus, anode half-cell is
${Zn}^{2 +} (aq) + 2 e^{-} ⟶ Zn (s); E^{\circ} = - 0 .76 V$
Cathode half-cell is ${Ag}_{2} O (s) + H_{2} O (l) + 2 e^{-} ⟶ 2 Ag (s) + 2 {OH}^{-} (aq)$
$\begin{array}{l} E^{\circ} = 0 .34 V \\ E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} \\ = 0 .34 - (- 0 .76) \\ = 1 .1 V \end{array}$

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