Question

Using data found in Appendix E of your textbook calculate the nonstandard emf for each of the following reactions if the concentration of each of the ions in these reactions is 0.0008 molar and everything else is standard (use 298 K for the temperature, R = 8.314 J/mol-K, and F = 96,485 C/mol):

(a) 1 Cl₂(g) + 2 Cr²⁺(aq) --> 2 Cl^-(aq) + 2 Cr³⁺(aq)   E = _____ V

(b) 1 Fe³⁺(aq) + 1 Na(s) --> 1 Fe²⁺(aq) + 1 Na⁺(aq)   E = _____ V

(c) 2 Li⁺(aq) + 2 I^-(aq) --> 2 Li(s) + 1 I₂(s)   E = _____ V

(d) 1 Mn²⁺(aq) + 1 Pb(s) --> 1 Mn(s) + 1 Pb²⁺(aq)   E = _____ V

Answer 1

(a) 1 Cl₂(g) + 2 Cr²⁺(aq) --> 2 Cl^-(aq) + 2 Cr³⁺(aq)  

anode reaction: oxidation takes place

Cr+2(aq) -------------------------> Cr+3 (aq) + e-   ,   E⁰_Cr+3/Cr+2 = - 0.41 V

cathode reaction : reduction takes palce

2 Cl-(aq) + 2e- -----------------------------> Cl2(g) , E⁰ = 1.36 V

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2Cr+2 + 2Cl- --------------------> 2Cr+3 + Cl2

E⁰_cell= E⁰_cathode- E⁰_anode

E⁰_cell= E⁰_cathode- E⁰_anode

         = 1.36 - (-0.41)

          = 1.77V

nernest equation

Ecell = E⁰_cell -2.303RT/nF* log [Cr+3]^2 /[Cr+2]^2 [Cl-]^2

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 =2

2.303RT/F= 0.0591

Ecell = E⁰_cell -(0.0591/n)* log[Cr+3]^2 /[Cr+2]^2 [Cl-]^2

Ecell = 1.77 - (0.059 x1/2) *log 1/ [0.0008]^2

Ecell   = 1.59 V

cell potential =Ecell   = 1.59 V

(b) similarly

Eo (cell) = 3.48 V

Ecell = E⁰_cell -(0.0591/n)* log [Na+] [Fe+2] / [Fe+3]

Ecell = 3.48 - (0.059 x1/1) *log   [0.0008]

Ecell   = 3.66 V