Question

By use of the following data at 25°C, solubility product(Ksp) of AgBr 4.88*10^-13 standard electrode potential: E(Ag/Ag)=0 .799 E(Br2(liquid)/Br) =1.065 V(1) Calculate the standard electrode potential for AgBr(s)/Ag at 25°C
(2) Design a cell for calculation of the standard Gibbs Energy formation of AgBr(s).

Answer 1

Answer on Question #53064 – Chemistry – General chemistry

Question:

By use of the following data at $25^{\circ}\mathrm{C}$, solubility product(Ksp) of AgBr 4.88*10^-13 standard electrode potential: E(Ag/Ag)=0.799 E(Br2(liquid)/Br) =1.065 V(1) Calculate the standard electrode potential for AgBr(s)/Ag at $25^{\circ}\mathrm{C}$

(2) Design a cell for calculation of the standard Gibbs Energy formation of AgBr(s).

Answer (1):

According to the conditions the cell should look like:

Ag|AgBr(s), HBr|Br₂ (pBr₂= 1 atm), Pt

The electrochemical process on the left side: Ag → Ag⁺ + e⁻

The electrochemical process on the right side: 1/2Br₂ + e⁻ → Br⁻

it means that the total reaction is: Ag + 1/2Br₂ → Ag⁺ + Br⁻ → AgBr

Using the Nernst equation for this process:

$$E_{\mathrm{AgBr/Ag, Br}^{-}} = E_{\mathrm{AgBr/Ag, Br}^{-}}^{0} - (\mathrm{RT/F}) \times \ln\left(\frac{C_{\mathrm{Ag}}, C_{\mathrm{Br}}}{P_{\mathrm{Br2}}}\right)$$

where C- the concentrations of species and pBr₂ - the pressures of bromine gas.

$$E_{\mathrm{AgBr/Ag, Br}^{-}} = E_{\mathrm{AgBr/Ag, Br}^{-}}^{0} - (\mathrm{RT/F}) \times \ln(K_{\mathrm{sp}} / p_{\mathrm{Br2}})$$ $$E_{\mathrm{AgBr/Ag, Br}^{-}}^{0} = E_{\mathrm{Ag+/Ag}}^{0} + E_{\mathrm{Br2/2Br}^{-}}^{0} = -0.799 \, \mathrm{V} + 1.065 \, \mathrm{V} = 0.266 \, \mathrm{V}$$

The standard potential at $25^{\circ}\mathrm{C}$ and the pressure 1 atm:

$$E_{\mathrm{AgBr/Ag, Br}^{-}} = 0.266 \, \mathrm{V} - 0.059 \times \lg K_{\mathrm{sp}} = 0.266 \, \mathrm{V} + 0.7264 \, \mathrm{V} = 0.9924 \, \mathrm{V}$$

Answer (2):

Scheme is the same:

Ag|AgBr(s), HBr|Br₂ (pBr₂= 1 atm), Pt

The standard Gibbs Energy formation of AgBr(s) at $25^{\circ}\mathrm{C}$ is calculated according to the equation:

$\Delta G_{T}^{0} = -F\Delta E$, where F - the Faraday number.

$\Delta G_{T}^{0} = -0.9924 \times 96493 \, \mathrm{J/mol} = -95759.65 \, \mathrm{J/mol} = -95,760 \, \mathrm{kJ/mol}$

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