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I've run into trouble on this question

Predict the standard potential at 310 K for the cell constructed from the hydrogen electrode and metal-insoluble salt electrode $\ce{Ag(s)|AgCl(s)|Cl-}$. Assume $\Delta_\mathrm rS$ is independent of $T$.

I know the reaction is $$\ce{AgCl +e- <=> Ag(s) + Cl- (aq)}$$ $$E^\circ=0.222$$

The left cell is a hydrogen electrode making the equation $$\ce{1/2H2 + AgCl(s) -> Ag(s) + Cl- (aq) + H+(aq)}$$ forming a nernst equation of

$$E=E^\circ-\frac{RT}{vF}\ln\left(\frac{\alpha_{\ce{Ag(s)}}\ \alpha_{\ce{Cl-}}\ \alpha{\ce{H+}}}{\alpha_{\ce{AgCl(s)}}\ \alpha_{\ce{H_2}}^{1/2}}\right)$$

which reduces to $$E=E^\circ-\frac{RT}{vF}\ln\left(\alpha_{\ce{Cl-}} \alpha_{\ce{H^{+}}}\right)$$

I know $R$, $T$, $v$ and $F$ but I don't understand where to begin to solve for $\alpha_{\ce{Cl-}} \text{and}\ \alpha_{\ce{H}}$

How would I find the activities?

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  • $\begingroup$ for a cell, there is two half reaction. You have the first one, you have now to find the reaction from the Hydrogen electrode side. Then combining both Nernst relation for both side maybe you'll make a step forward. $\endgroup$ – Babounet Dec 9 '14 at 18:22
  • $\begingroup$ If you notice my Nernst in the question includes the activity of H. $\endgroup$ – John Snow Dec 9 '14 at 18:25
  • $\begingroup$ Yeah, that's why I told you that writing the Nernst equation from the hydrogen electrode reaction could help, like this, you can combine both, and guess out what is this alpha bothering you. $\endgroup$ – Babounet Dec 9 '14 at 18:27
  • $\begingroup$ I'm sorry for the confusion, what i meant to imply is that the nernst i displayed is derived from including the hydrogen half reaction. $\endgroup$ – John Snow Dec 9 '14 at 18:29
  • $\begingroup$ Oh ok, so you should know what is this alpha then ? $\endgroup$ – Babounet Dec 9 '14 at 18:30
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Predict the standard potential at 310 K for the cell constructed from the hydrogen electrode and metal-insoluble salt electrode $\ce{Ag(s)|AgCl(s)|Cl−}$

The reactions would be: $$\ce{AgCl +e- ->Ag +Cl-}\tag{$E_1$ V}$$ $$\ce{2H+ +e- ->H2}\tag{0 V}$$ Adding: $$\ce{AgCl +H_2->Ag +Cl- +2H+}\tag{$E_1 V$}$$

The standard potential would be: $$E=E^\circ -\frac{RT}{\nu F}\log\ce{\frac{[Cl- ][H+]}{P_{H2}^2}}\quad (E^\circ =E_1,\nu=1)$$ H Now you have $E_1$, also you must have $\ce{[Cl- ],[H+ ]}$ beforehand, there's no other way, other than considering the fact that (standard) Hydrogen electrode is maintained at $\text{1M}$ and $\text{1 atm}$, and to maintain charge $\ce{[Cl- ] =[H+ ]}$.In that case, $$E=0.222-\frac{8.314\times 310}{96500}\ln(1)=0.222$$


Just for info (from Wikipedia): $$\begin{array}{c|c}\text{Temperature}&\text{Potential} E^\circ\\ ^\circ\text{K} &\text{V versus SHE at the same temperature}\\\hline 298 &0.22233\\ 333 &0.1968\\ 398 &0.1330\\ 423 &0.1032\\ \end{array}$$ Also when you consider $\Delta S\ne\text{ constant}$:(A.J. Bard, R. Parson, J. Jordan, "Standard Potentials in Aqueous Solution", Marcel Dekker, Inc., 1985.) $$E^\circ=0.23695 - 4.8564\times10^{−4}t - 3.4205\times10^{−6}t^2 - 5.869 \times 10^{−9}t^3$$

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  • $\begingroup$ @JohnSnow where did you find the question, did you had more info, have any answer to this? $\endgroup$ – RE60K Dec 10 '14 at 17:54
  • $\begingroup$ The question was given to me by my physical chemistry professor. the only other statement was " assume $\Delta_r S$ is independent of T." $\endgroup$ – John Snow Dec 10 '14 at 18:03
  • $\begingroup$ you asked in your edit "how can I find the activities" but I already answered to this in the caht you oppened ... so I'll say it again, for species in solution, the activity is the concentration time the so called gamma coefficient (but often approximated to be one in practical case, either way those coefficients are tabulated) and for a gaz, the activity is the partial pressure. Since you've already been told that for a NHE (Normal Hydrogen Electrode) you can assume the concentration of H+ to be 1M and the partial pressure of H2 to be 1 atm, you can answer the question now. $\endgroup$ – Babounet Dec 10 '14 at 18:08
  • $\begingroup$ @babounet if they are both equal to one then ln1= 0 canceling out the rt/vf term. $\endgroup$ – John Snow Dec 10 '14 at 18:31
  • $\begingroup$ and ? isn't it an acceptable result to have E = E° ? $\endgroup$ – Babounet Dec 10 '14 at 18:41
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Here's the answer from the key:

$$\begin{align}{2}\frac{1}{2}\ce{H2 &-> H+}+e^- \tag{0V}\\ \\ \ce{AgCl(s) +e^- &-> Ag(s) +Cl-} \tag{0.22V}\\\\ &E^o =0.22V\ \text{at 298 K} \\\\ \ce{2AgCl+H2 &->Ag(s)+2HCl(aq)} \\\\ \Delta_rS &= \frac{vF(E^0_1-E^o_2)}{T_1-T_2} \\\\ \Delta_rS &=S^o_m(\ce{Ag(s)}) +S^o_m(\ce{2HCl(aq)}) - S^o_m(\ce{H2}) -S^o_m(\ce{2AgCl})\\ &=(2\cdot 42.55+2\cdot 56.5-130.68-2\cdot 96.2) \\ &= -124.98 \\ -124.98 &= \frac{1 \cdot 9.6485\ x \ 10^4 (0.22-E^o)}{298-310} \\ 0.22-\frac{1499.66}{9.6485x10^4} &= E^o \\\\ E^o_{(310)} &= 0.204V \end{align}$$

Thanks anyway for the attempts

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