Question

# What is the cell potential for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 47 ∘C when [Fe2+]= 3.10 M...

What is the cell potential for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 47 ∘C when [Fe2+]= 3.10 M and [Mg2+]= 0.110 MM . Express the potential numerically in volts.

We need to use the following equation:

Ecell = Eº - RT/nF lnQ

Where R = 8.314,

T = 47 °C = 320.15 K

F = 96485

So we have to find E°, n and ln(Q) first.

Step 1: Finding E° and n

Write the semi reactions and read the E° in a potentials table:

Mg (s)   ----> Mg+2 + 2e-      E° = 2.356 V

Fe+2 + 2e- ----> Fe (s)        E° = -0.440

_______________________________

Mg (s) + Fe+2 ----> Mg+2 + Fe (s)   E° = 1.676 - 0.440 = 1.92 V

n is simply the number of trading electrons which is 2.

Step 2: Find Q

Q is defined by:

Q = [products]/[reagents] = 0.110 M / 3.10 M = 0.0355

Step 4: Replace values

Ecell = Eº - RT/nF lnQ = 1.92 V - [ (8.314)*(320.15) / ((2) * (96485))] * ln(0.0355) = 1.97 V

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