Please show all work for question 79 including all formula\'s, constants, and co

Question

Please show all work for question 79 including all formula's, constants, and conversions. Thank you! The answers are a.) 0.23 and b.) 1.2*10^-5

C Consider The Concentr dCI ch13 to. riodic table Untitled C Secure I h ps:/vi 148 277 of 284 Learn 882 Chapter 18 Electrochemistry a. Determine the overall cell reaction and calculate E a. Calculate the concentration of Ag at the anode b. Calculate AG and K for the cc reaction at 25°C b. Dctcrminc thc valuc of thc cquilibrium constant for thc formation of Ag(S203) 3+ c. Calculate at 25°C when [Au 0 X 10 2 M and 0 X 10 4 M Ag (aq) 2S20,2 (aq) Ag(S203 2 (aq 79 An electrochemical cell consists of a standard hydrogen clec Under standard conditions, what reaction occurs, if any, when trode and a copper metal electrode. cach of thc following opcrations is pcrformcd? a. What is thc potential of thc ccll at 25 C if thc coppcr a. Crystals of are added to a solution of NaCl electrode is placed in a solution in which LCu b. Cl gas is bubblcd into a solution of Nal 2.5 X10 4 M? c. A silver wire is placed in a solution of CuCl2 b. The copper electrode is placed in a solution of unknown d. An acidic solution of FeSO is exposed to air Cuz J. The measured potential at 25°C is 0.195 V. What is ICu +1? (Assume Cu 2+ s reduced For thc rcactions that occur. writc a balanccd cquation and e E AGO, and K at 25°C. 80. An electrochemical cell consists of a nickel metal electrode calcul mmersed in a solution with IN 1 0 M scparatcd by a 86. A disproportionation reaction involvcs a substance that acts as porous disk from an aluminum metal electrode. both an oxidizing and a reducing agent, producing higher and ower oxidation states of the same elemen n the products. a. What is the potential of this cell at 25°C if the aluminum Which of the following disproportionation reactions are spon electrode is placed in a solution in which LAl 7.2 X taneous under standard conditions? Calculate AG and K at 10 3 M? 25 C for thosc rcactions that arc spontancous undcr standard b. When the aluminum clcctrod is placcd in a ccrtain solu c condi ns. tion in which IAI Jis unknown, the measured cell poten a. 2Cu aq) Cu aq) Cu tial at 25°C is 1.62 V. Calculate [A n the unknown solution. (Assume Al is oxidized b. 3Fc2 2Fe3+ Fc (s) Ask me anything 618 PM 4/18/2017 3

Explanation / Answer

a) The half cell reactions are:

Cu2+ (aq) + 2 e- ------> Cu (s); E0 = +0.337 V

2 H+ (aq) + 2 e- -----> H2 (g); E0 = 0.000 V

The more positive the standard electrode potential, the more is the tendency of the element to be reduced. Therefore, Cu2+ will be reduced while H2 will be oxidized. The electrode reactions are:

Reduction: Cu2+ (aq) + 2 e- ------> Cu (s); E0red = +0.337 V

Oxidation: H2 (g) -----> 2 H+ (aq) + 2 e-; E0ox = 0.000 V

The overall cell reaction is

Cu2+ (aq) + H2 (g) ------> Cu (s) + 2 H+ (aq)

E0cell = E0red + E0ox = (+0.337 V) + (0.000 V) = 0.337 V

Assuming standard conditions for H2 and H+ (i.e, 1 atm pressure and 1 M solution) with [Cu2+] = 2.5*10-4 M, we have

Ecell = E0cell - (2.302*R*T/n*F)*log (1/[Cu2+]) …..(1)

where R = 8.314 J/mol.K; T = 298 K (25C), n = number of moles of electrons transferred = 2 mole and F = 1 Faraday of electricity = 96500 C = 96485 J/V.

Plug in values and write

Ecell = (0.337 V) - [2.302*(8.314 J/mol.K)*(298 K)/(2 mole).(96500 J/V)]*log[1/(2.5*10-4)] = (0.337 V) – (0.02955 V)*log (4000) = (0.337 V) – (0.02955 V)*(3.602) = (0.337 V) – (0.1064 V) = 0.2306 V 0.23 V (ans)

b) Use expression (1) from above and put Ecell = 0.195 V

0.195 V = (0.337 V) – (0.02955 V)*log (1/[Cu2+])

===> (-0.02955 V)*log (1/[Cu2+]) = -0.142 V

===> log (1/[Cu2+]) = (0.142 V)/(0.02955) = 4.805

===> 1/[Cu2+] = antilog (4.805) = 63826.35

===> [Cu2+] = 1/63826.35 = 1.566*10-5 1.6*10-5

The concentration of Cu2+ = 1.6*10-5 M (ans).

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