Equation help needed, 100% lost on both parts of this problem! Help! Will rate w

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Equation help needed, 100% lost on both parts of this problem! Help! Will rate with awesome review for help and explanation and correct answer for this problem I have literally been working on it for hours! THanks! I need help writing the net ionic equation and then from that calculating E.

Silverware Tamish Low concentrations of hydrogen sulfide in air react with silver to form Ag2S, more familiar to us as tarnish. Silver polish contains aluminum metal powder in an alkaline suspension. Write a balanced net ionic equation for the redox reaction of Ag2S and Al metal that produces Ag metal and AI(OH)3. Be sure to include phases for each species in the equation. Calculate E degree for the reaction in Part a. [Hint: Derive E degree values for the half-reactions in which Ag2S is reduced to Ag metal and AI(OH)3 is reduced to Al metal. You can do this by combining the standard potentials for the reduction of Ag+ and Al3+ in Appendix 6 with the Ksp expressions and values of Ag2S(Ksp = 1.6x10-49) and AI(OH)3 (Ksp = 1.9 times 10-33).

Explanation / Answer

a) solution

anode :    oxidation

Al(s) ------------------- Al+3 (aq) +3e-

cathode :    reduction

2Ag + (aq) + 2e-   --------------------> Ag (s)

balanced net reaction:

2Al(s)+    6Ag + (aq)   ------------------- 2Al+3 (aq) + 6Ag(s)

b) solution

E0cell= E0cathode- E0anode

E0cell= E0Ag+/Ag- E0Al+3/Al

          = 0.80 - (-1.66)

        E0cell = 2.46 V

now from Ksp values calculation of concentration of Ag+ and Al+3

Ag2S   --------------------> 2 Ag+   + S^-2

                                      2S            S

Ksp = (2S)^2(S) = 4S^3

1.6 x 10^-49 = 4S^3

S= 3.42 x 10^-17 M

[Ag+] = 2 x S = 2x3.42 x 10^-17

[Ag+] = 6.84 x 10^-17 M --------------------------------------------(1)

now calculating Al+3 concentration

Al(OH)3 -----------------------------> Al+3 + 3OH-

                                                     S          3S

Ksp = [Al+3][OH-]^3

        = (S)(3S)^3

Ksp      = 27 S^4

1.9 x 10^-33 = 27 S^4

S= 2.9 x 10^-9

[Al+3] = S = 2.9 x 10^-9 M ---------------------------------------------(2)

we use nernest equation

Ecell = E0cell -RT/nF* log {[Al+3]^2 / [Ag+]^6}

Here R= universal gas constant 8.314 J/K mol

T = absolute temperature =25(0C)= 298k

F= faraday = 96500 Coloumb/mol

     n   = no of moles of electrons are transfered =6

RT/F= 0.0591

Ecell = E0cell - (0.0591/n)* log {[ 2.9 x 10^-9]^2 / [Ag+]^6}

        = 2.46 - (0.0591/6)* log {[Al+3]^2 / [6.84 x 10^-17]^6}

        = 1.672

Ecell= 1.672 V

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