Question

Calculate the cell potential for the following reaction as written at 25.00degreeC. given that (Cr^2+] = 0.849 M and [Sn^2+] = 0.0130 M. Standard reduction potentials can be found here.

Answer 1

anode reaction: oxidation takes place

Cr(s) -------------------------> Cr+2 (aq) + 2e-   ,   E⁰_Cr+2/Cr = - 0.91 V

cathode reaction : reduction takes palce

Sn+2(aq) + 2e- -----------------------------> Sn(s) , E⁰_Sn+2/Sn = -0.14 V

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net reaction: Cr(s) +Sn+2(aq) -------------------------> Cr+2 (aq) + Sn(s)

E⁰_cell= E⁰_cathode- E⁰_anode

E⁰_cell = - 0.14 - (-0.91)

          = 0.77 V

nernest equation

Ecell = E⁰_cell -2.303RT/nF* log [Cr+2]/[Sn+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+2]/[Sn+2]

Ecell = 0.77 - (0.059 x1/2) *log 0.849/0.0130}

Ecell   = 0.72 V

cell potential =Ecell   = 0.268 V