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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, 2014 at 18:22
  • $\begingroup$ If you notice my Nernst in the question includes the activity of H. $\endgroup$
    – John Snow
    Dec 9, 2014 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, 2014 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, 2014 at 18:29
  • $\begingroup$ Oh ok, so you should know what is this alpha then ? $\endgroup$
    – Babounet
    Dec 9, 2014 at 18:30

2 Answers 2

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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, 2014 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, 2014 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, 2014 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, 2014 at 18:31
  • $\begingroup$ and ? isn't it an acceptable result to have E = E° ? $\endgroup$
    – Babounet
    Dec 10, 2014 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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  • $\begingroup$ There is a slight mistake to the answer key given, John. In the last step while putting in the value of v, you should put 2 instead of 1, as all calculations of entropy change were done taken into consideration of 2 mole electron transfer. So ans should rather be 0.212 approx $\endgroup$ Dec 30, 2022 at 16:41

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