Consider the following E o values E F e 3 + / F e 2 + o = 0.77 V E S n 2 + / S n o = - 0.14 V Under standard conditions the potential for the reaction S n ( s ) + 2 F e 3 + ( a q ) → 2 F e 2 + ( a q ) + S n 2 + ( a q )

Step-by-step solution

S n ( s ) + 2 F e 3 + ( a q ) → 2 F e 2 + ( a q ) + S n 2 + ( a q ) E cell o = E reduction o cathode − E reduction o anode E cell o = 0.77 − ( − 0.14 ) ∴   E cell o = − 0.91   V