Calculate delta G for the following rxn at 25 degrees celcius. 3Mn2+ (aq) + 2Al(s) -><- 3Mn(s) + 2Al3+ (aq)

The
anode= Mn2+ + 2e- -> Mn
= -1.18 V
The
Cathode= Al3+ + 3e- -> Al
= -1.66 V

The Ecell= -1.66-(-1.18)= -0.48 V

But I do not know how to find Delta G.
Delta G= -nFE

The answer is in kj/mol.
Can anyone help?

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