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I came across a question today.

A galvanic cell is composed of two hydrogen electrodes, one of which is a standard one. In which of the following solutions should the other electrode be immersed to get maximum e.m.f. ($K_a=10^{-5}$ for acetic acid and $k_a=10^{-3}$ for phosphoric acid)

(A) $0.1$ $M$ $\ce{HCl}$
(B) $0.1$ $M$ $\ce{CH3COOH}$
(C) $0.1$ $M$ $\ce{H3PO3}$
(D) $0.1$ $M$ $\ce{H2SO4}$

I thought "Well, seems like all I need to do is find which solution will give the most number of hydrogen ions because its hydrogen ions which will suck those electrons from the other half cell which will be the cause of current. And concentration is same for all. So there is no chance for a weak acid. So option (B) cannot be correct. All other seems like a strong acid. So it should be $\ce{H2SO4}$ as it gives $0.2$ $M$ of hydrogen ions for $0.1$ $M$."

But my thought was absolutely wrong. Why?

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  • $\begingroup$ There are more hydrogen ions on the other side, anyway. $\endgroup$ – Ivan Neretin Feb 2 '16 at 6:34
  • $\begingroup$ @IvanNeretin you mean on the standard half cell? But why is it? $\endgroup$ – manshu Feb 2 '16 at 7:39
  • $\begingroup$ Yes, that half. See, what is the standard hydrogen electrode? What is the concentration of $\ce H^+$ in it? $\endgroup$ – Ivan Neretin Feb 2 '16 at 7:43
  • $\begingroup$ @IvanNeretin its unity. So do you mean $\log\ce{H+}-\log {\ce{SO4^2-}} = 10$ while if there was $\ce{CH3COO-}$ it would have been lot more...is that what you mean? $\endgroup$ – manshu Feb 2 '16 at 7:59
  • $\begingroup$ I don't see where that 10 comes from. Other than that, yes. The SHE is more acidic than any of our options. If we want to be as different from it as possible, we want to be less acidic. $\endgroup$ – Ivan Neretin Feb 2 '16 at 8:02
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This is a concentration cell question where 2 electrodes of same stuff are dipped in their own ions and if there is non equal concentrations of anything a "potential" develops called electrode potential (to be understood in qualitative way)(kind of more concentration of reactant forces product formation) as you can understand this would be a redox reaction where reduction would occur in one and oxidation in another, Now, by standard it means that all the components of the reaction (participants, i.e. reactants and products) are in standard state i.e. $1M, 1$ $bar$ etc, so, we know that concentration of $\ce{H+}$ in standard one is $1M$, we need to create biggest gap between them, as the concentration is really high and we cant compete with it, just reduce our own $\ce{H+}$ to increase their difference and force ours to undergo Oxidation to gas and standard undergoes reduction

Just for extra information, $E=E_0-{RT\over nF}\ln Q$ and can be derived by $G=G_0+RT\ln Q,\;G=-nFE$ {$n$ is number of electrons being exchanged ex- in $\ce{2H+->H2},\;n=2$; $F$ is the charge on 1 mol electrons} and Standard Hydrogen is the standard bearer so, we assign its standard value as $0$, $E_0=0$.

EDIT : I can't comment due to reputation , basically, assume complete dissociation and and concentration of $\ce{H+}$ in $\ce{HCl}$ would be $0.1M$ , $\ce{H2SO4}$'s second ionization is not complete like others but can be assumed for a good approximation, so concentration becomes $0.2 M$ , and basic thing to remember is

IF ONE SIDE UNDERGOES OXIDATION THEN OTHER MUST UNDERGO REDUCTION

and if we strengthen either reaction, potential increases, concentration is only in a solvent, how can a solid would have active $\ce{H+}$ ions (everything is activity not concentration or partial pressure flat out, they are approximated, solids are assumed to have activity as unity) and log is defined as the power to $10$ that would give us that number

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