The emf generated by the cell when [Al3+] = 3.5 × 10-3 M and [I−] = 0.30 M is ________ V.?
The standard emf for the cell using the overall cell reaction below is +2.20 V:
2Al(s)+3I2(s)→2Al3+(aq)+6I−(aq)
The emf generated by the cell when [Al3+] = 3.5 × 10-3 M and [I−] = 0.30 M is ________ V.
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Nernst Equation.
Ecell = emf of cell
E0 = standard cell potential
n = electrons transferred
T = 298 k
R = 8.314 J mol–1K–1 and F = 96,485 J V–1 mol–1),
Q is reaction quotient
If we insert these values in Nerst equation
Ecell = E0 cell - 0.0592 V/n * log Q where Q = [ Al+3 ]^2 * [ I -]^6 / [ Al ]^2 * [ I2 ] ^3
Activity of solids Al and I 2 is 1.
[ Al+3 ]^2 * [ I -]^6 = 1 ( denominator above is 1 )
or Q = [ 3.5 * 10 -3 ]^2 * [0.3 ]^6
n = 6 as 6 moles of electrons are transferred from 2 moles Al to 3 moles I2
2Al(s) ------> 2Al3+ (aq) + 6e
3I2(s) + 6e ------> 6I- (aq)
E cell = 2.2 - 0.0592/6*Log (3.5*10-3)^2*(0.30)^6
this equals to 2.2 - 0.0592/6 *(-0.12697)
or, E cell = [ 2.2 - (-0.00125)] = 2.201 V