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

To calculate Ksp for AgBr, use the given standard reduction potentials and the Nernst equation. The equilibrium constant, K, for the half-reaction can be found using the Nernst equation. The equilibrium constant Ksp can then be calculated from K.

Step-by-step explanation:

To calculate Ksp for AgBr, we can use the given standard reduction potentials and the Nernst equation. The balanced half-reaction for the reduction of AgBr(s) to Ag(s) is:

AgBr(s) + e- → Ag(s) + Br-(aq)

The reduction potential for this half-reaction is +0.071 V. Using the Nernst equation, we can find the equilibrium constant, K, for this reaction:

K = [Ag(s)][Br-(aq)] / [AgBr(s)] = 10^(nE°/0.0592) = 10^((1)(0.071)/0.0592) = 1.44

Since the stoichiometry of the reaction is 1:1, the equilibrium constant Ksp for AgBr can be directly calculated from K:

Ksp = [Ag+(aq)][Br-(aq)] = 1.441

Answer 2

`k_(sp)=4.7 * 10^(-13).`

Step-by-step 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* F * E^o= -R* T* ln(k_s_p)\\-1* 96485* (-0.729)=-8.314* 298* ln(k_s_p)\\k_s_p=4.7* 10^(-13).`

`k_(sp)=4.7 * 10^(-13).`

Hence, this is the required solution.