Chemistry reduction potential question?

Question

Consider two hypothetical metals, X and Y which can exist as ions X(2+) and Y(3+) respectively. For these metals, the standard reduction potentials are:

X(2+)+2e(-) --> X -2.276V

Y(3+)+3e(-) --> Y 2.154V

Compare the following Voltaic cells:

X(s) | X(2+) (.305M) || Y(3+) (.231M) | Y(s), Ecell

X(s) | X(2+) (1M) || Y(3+) (1M) | Y(s), E^0cell

At 25deg C, Ecell-E^0cell=??

Answer 1

X(2+)+2e(-) --> X     -2.276V  (reduction)

Y(3+)+3e(-) --> Y    2.154V   (reduction)

voltaic cell:

X  --> X(2+)+2e(-) -->   +2.276V  (oxidation)

Y(3+)+3e(-) --> Y    2.154V   (reduction)

2.276 - 2.154 = 0.122 volts

first answer : Eo = + 0.122 volts

==========================

to balance the electrons lost & taken we triple the X rxn & double the Y reaction:

3X  --> 3X(2+)+6e(-) -->   +2.276V  (oxidation)

2Y(3+)+6e(-) -->2 Y    2.154V   (reduction)

2Y+3  & 3X -->  3 X+2  & 2 Y

Q = [X+2]^3  / [Y+3]^2

=====================================

nernst

E = Eo - (0.0592 / n ) (Log Q)

E = 0.122 volts  - (0.0592 / 6 electrons ) (Log prod / reactants)

E = 0.122 volts  - (0.00987 ) (Log [X+2]^3 / [Y+3]^2 )

E = 0.122 volts  - (0.00987 ) (Log [0.305]^3 / [0.231]^2 )

E = 0.122 volts  - (0.00987 ) (Log 0.0284 / 0.0534)

E = 0.122 volts  - (0.00987 ) (Log 0.532)

E = 0.122 volts  - (0.00987 ) (- 0.274)

E = 0.122 volts  - (- 0.00271)

E = + 0.1247

you last answer E = + 0.125 volts