Question

The value of E° for the following reaction is 0.34 V. What will the cell potential be when the [Cd2+] = 10.0 M and the [Cr3+] = 1.0 M?

Answer 1

anode reaction: oxidation takes place

Cr(s) -------------------------> Cr+3 (aq) + 3e-   ,   E⁰_Cr+3/Cr = - 0.74 V

cathode reaction : reduction takes palce

Cd+2 + 2e- -----------------------------> Cd(s) , E⁰_Cd+2/Cd = -0.40V

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net reaction: Cr(s) +Cd+2(aq) -------------------------> Cr+3 (aq) + Cd(s)

nernest equation

Ecell = E⁰_cell -2.303RT/nF* log [Cr+3]/[Cd+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 = 6

2.303RT/F= 0.0591

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

Ecell = 0.34 - (0.059 /6) *log 1.0/10.0}

Ecell   = 0.35 V

cell potential =Ecell   = 0.35 V