Question

Ag+(aq) + e- → Ag(s) E° = +0.800 V AgBr(s) + e- → Ag(s) + Br-(aq) E° = +0.071 V Br2(l) + 2 e- → 2 Br-(aq) E° = +1.066 V Use some of the data above to calculate Ksp at 25°C for AgBr. Enter your answer in exponential format (sample 1.23E-4) with two decimal places and no units.

Answer 1

Answer:

$k_{sp}=4.7 \times 10^{-13}.$

Explanation:

Now equation of tqo halves are:

Oxidation : $Ag(s)-->Ag^+(aq)+e^-$

Reduction : $AgBr(s)+e^--->Ag(s)+Br^-(aq)$

We know,

$E^o_{cell}=0.071-(0.8)=-0.729\ V.$

$\Delta G^o=-n\times F \times E^o= -R\times T\times ln(k_s_p)\\-1\times 96485\times (-0.729)=-8.314\times 298\times ln(k_s_p)\\k_s_p=4.7\times 10^{-13}.$

$k_{sp}=4.7 \times 10^{-13}.$

Hence, this is the required solution.