1. Tardigrade
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  3. Chemistry
  4. The standard reduction potential for two reactions is given below. AgCl(s) +e- → Ag(s) +Cl-(aq); E° =0.22V Ag+(aq) +e- → Ag(s); E° = 0.08V The solubility product of AgCl under standard conditions of temperature is given by

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Q. The standard reduction potential for two reactions is given below.
$AgCl_{\left(s\right)} +e^{-} \to Ag_{\left(s\right)} +Cl^{-}_{\left(aq\right)}; E^{\circ} =0.22V$
$Ag^{+}_{\left(aq\right)} +e^{-} \to Ag_{\left(s\right)}; E^{\circ} = 0.08V$
The solubility product of $AgCl$ under standard conditions of temperature is given by

Electrochemistry

Solution:

$AgCl \rightleftharpoons Ag^{+} + Cl^{-} K_{sp} = \left[Ag^{+}\right] \left[Cl^{-}\right]$
$\frac{AgCl_{\left(s\right)} +e^{-} \to Ag_{\left(s\right)} +Cl^{-}_{\left(aq\right)}; E^{\circ} =0.22V Ag^{+}_{\left(aq\right)} +e^{-} \to Ag_{\left(s\right)}; E^{\circ} = 0.08V}{AgCl \to Ag^{+}+Cl^{-}; E^{\circ}_{cell} =0.22 +\left(-0.80\right) = -0.58V}$
Applying $E^{\circ}_{cell} =\frac{0.0591}{n}$ lof $K_{sp}$
log $K_{sp} = - \frac{0.58 \times 1}{0.0591} = -9.8138$
Taking antilog $K_{sp} = 1.5 \times 10^{-10}$