24. Calculate Eocea for a silver-aluminum cell in which the cell reaction is A)

Question

24. Calculate Eocea for a silver-aluminum cell in which the cell reaction is A) B) C) D) E) Alis) + 3Ag+(aq) All"(aq) + 3Ag(s) -2.46 V 0.86 V -0.86 V 2.46 V none of these 25. Consider the following standard reduction potentials in acid solution: AP+ + 3e- Al(s) +0.07 +0.14 +0.77 AgBr(s) + e- Ag(s) + Br Sn4+ + 2e- Sn2. Fe3+ + e- Fe2+ strongest oxidizing agent among those shown above is B) Fe C) Br D) Al+. E) Al. 20 For the electrochemical cell, cd(s)| Ca'(ag) Co (aq)| Cofs), determine the equilibrium constant (Ke) at 25 C for the reaction that occurs. A) 1.1 x 10 B) 1.1 x 1023 C) 9.4x 1022 D) 1.1 x 10 E) 5.8 x 10 27, Calculate Go for the electrochemical cell Fe(s) l Fe2(aq! Sn4+(aq) I Sn."(aq) I Pus) A) -5.5 x 10 kJ/mol B) -1.1 x 10? kJ/mol C) 6.0x 10' kJ/mol D) 1.1 x 103 kJ/mol E) 1.2 x 10 kJ/mol

Explanation / Answer

24. E0cell = Ecathode- Eanode    ( reduction potential of cathode > anode always)

           = 0.8 - -1.662

       = 2.464 V

25. oxidizing agent : the substance or ion which undergoes oxidation easily.

the element with lowest E0red, could readily participates in oxidation reaction.

answer: E

26. E0cell = -0.28 - (-0.4)

           = 0.12 V

dG0 = -nFE0cell

     = -2*96500*0.12

    = -2316o joule.

DG0 = -RTlnK

-23160 = -8.314*298lnk

k = 1.14*10^4

27)

E0cell = 0.15- -0.44 = 0.59 V

    DG0 = -nFE0cell

        = -2*96500*0.59

    = -113.87 kj/mol

answer: B

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