Consider the cell....

Question

In I In^3 (1M) II Ru^3 (.1M) I Ru^2 (1M)

the cell voltage is

-less than E^0

-greater than E^0

-the same as E^0

or not enough info

Answer 1

We know the overall equation of the cell is

ln(s) --> In^3+(aq) + 3e-

Ru^3+(aq) + 3e- --> Ru(s)

_______________________

In(s) + Ru^3+(aq) --> In^3+(aq) + Ru(s)

Since equilibrium quotient, Q = [In^3+]/[Ru^3+], then

Q = 1/0.1 = 10

ΔE = ΔEo - (RT/nF)lnQ

ln10 = 2.30258

Thus, ΔE must be LESS than ΔEo.

[Answer: see above]

Answer 2

NOT ENOUGH INFO

because with all sign conventions,when u talk about "E"

and "E0" u lose the sign.THE CELL VOLTAGE ALLWAYS IS

A POSITIVE NUMBER(U NEVER SAY E=(-3V)<E0=(-2V)  ).if not convinced,try to put side by side 2 such cells,one with standard cond.:In/In3+(1M)//Ru3+(1M)/Ru2+(1M).the... close the circuit.Le Chatelier's Principle will bring up a contradiction>passage of a current,free,increases the difference.there are 4 cases:2 polarities,indium or ruthenium positive,and 2 in-equalities,E>E0 E<E0.

caution:the above klingonian used some equation in some fashion to obtain some result and for some reason he likes some conclusion.

did i say "KLINGONIAN"? I MENT http://www.urbandictionary.com/define.ph...