[Fe3+]= 2.80 M ; [Mg2+]= 2.1×10−3 M. Express your answer in units of volts. E0 =...

[Fe3+]= 2.80 M ; [Mg2+]= 2.1×10⁻³ M. Express your answer in units of volts.

E0 = 2.33

Expert Solution

3Mg(s) -------------------> 3Mg^2+ (aq) + 6e^- E0 = 2.38V

2Fe^3+ (aq) + 6e^- ---------> 2Fe(s) E0 = -0.04v

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3Mg(s) +2Fe^3+(aq) --------> 3Mg^2+(aq) + Fe(s) E0cell = 2.33v

n = 6

Ecell = E0cell + 0.0592/n log[Mg^2+]^3/[Fe^3+]^2

= 2.33 + 0.0592/6 log(2.1*10^-3)^3/(2.8)^2

= 2.33 +0.00986log1.18125*10^-9

= 2.33+0.00986*-8.9276 = 2.2419v >>>>answer

queen_honey_blossom answered 2 years ago

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