Chemistry, 27.05.2020 18:00 zachzach28280

A student measures Ecell of a concentration cell created by having the saturated solution of AgBr on one side and 0.10 M Ag+(aq) on the other (along with silver electrodes). The experimental value is 0.603 V. The student is told to use the theoretical value of the Nernst plot (−0.0591 V). (The concentration of the halide is also 0.10 M.) (a) In which direction will electrons spontaneously flow? from the 'saturated side' to the '0.10 M side' from the '0.10 M side' to the 'saturated side' the electrons will not flow in this cell (b) What is the value of [Ag+]dilute? M (c) What is the experimental Ksp? (d) Calculate ΔG° in kJ/mol using ΔG° = −RTlnKsp. (R = 8.314 J/K · mol, T = 298 K.) kJ/mol

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Answer from: petriajack5543

Complete Question

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From the saturated side to the '0.10 M ' side

The diluted Ag^+ is $[Ag^+] = 4.45*10^{-12} \ M$

The experimental values of $k_{sp}$ is $K_{sp} = 4.45 *10^{-13}$

$\Delta G ^o = 70.478 \ K J /mol$

Explanation:

From the question we are told that

The concentration of $Ag^+ _{(aq)}$ is $[Ag_{(aq)}] = 0.10 M$

The experiment value is E_e = 0.603 V

The value of the Nernst plot is $E_n = - 0.0591 \ V$

The concentration of the halide is [Br] = 0.10 M

From the question we are told that one side has a saturated AgBr and the other side has a ion 0.10M Ag+(aq) this will cause a difference in potential causing the electron to flow from the the saturated side to the side with Ag+(aq) in other to reduce it

The equation for the reaction is given as.

$AgBr_{aq} + e^- \to Ag_{s} + Br^-_{aq}$ E^o = 0.07 V

$AgBr_{aq} + e^- \to Ag_{s} + Br^-_{aq}$ E^o = 0.80 V

This values are constant potential data for standard reduction of AgBr

The overall equation of reaction in the cell is

$Ag ^+ + Br^- \to AgBr _{(aq)}$

$E_{cell} = 0.80 - 0.07$

$E_{cell} = 0.73 V$

This potential for the cell can also be represented as

$E_{cell} = \frac{0.0591}{n_e} log \frac{1}{k_{sp} }$

Where is the number of moles of electron which 1 from the chemical equation

$k_{sp}$ is the solubility product which is mathematically represented as

$k_{sp} = [Ag^+][Br^-]$

$0.73 = \frac{0.0591}{n_e} log \frac{1}{k_{sp} }$

=> $log \frac{1}{K_{sp}} = \frac{0.73}{0.0591}$

=> $K_{sp} = 4.45 *10^{-13}$

Now

$k_{sp} = [Ag^+][Br^-]$

$[Ag^+] = \frac{k_{sp} }{[Br^-]}$

$[Ag^+] = \frac{4.45*10^{-13} }{0.10}$

$[Ag^+] = 4.45*10^{-12} \ M$

The Free energy ΔG° is mathematically represented as

$\Delta G ^o = RTln k_{sp}$

Where R is the gas constant with value $R = 8.314 J/K \cdot mol$

$\Delta G ^o = -8.314 * 298 * ln (4.45*10^{-13})$

$\Delta G ^o = 70478.3 J /mol$

Converting to KJ

$\Delta G ^o = 70.478 \ K J /mol$

A student measures Ecell of a concentration cell created by having the saturated solution of AgBr on

Answer from: Quest

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

Answer from: Quest

Ithink it is quantity of plants

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A student measures Ecell of a concentration cell created by having the saturated solution of AgBr on...