Consider a galvanic cell composed of two \(\text{H}^{+}\left|\right.\text{H}_2\)-half cells where \( \left[\text{H}^+\right] \) are \(0.025 \text{M} \) and \( 5 \text{M} \) in the first and second half-cell, respectively. Calculate the electromotive force (EMF).
It's good practice to first visualize such a problem before attempting to solve it in order to gain a clear understanding of the system. The galvanic cell could look like this:
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| Concentration chain of two hydrogen electrodes. The blue arc indicates the salt-bridge. The hydrogen pressures do not need to be the same (see text). |
A galvanic cell with two times the same system at different electrolyte concentration is refered to as concentration chain. At the platinum electrode, the hydrogen gas is oxidized and \(\text{H}^+\) ions enter into solution. A thermodynamic equilibrium is reached when \( \left[\text{H}^+\right] \) is equal in both cells.
We can recognize that in electrode 1, \( \left[\text{H}^+\right] \) is lower and so it is in electrode 1 where hydrogen is oxidized. By convention, the electrode where oxidation occurs is referred to as anode, the one where reduction occurs as kathode. The electrons thus exit the cell at electrode 1, run through the wire to electrode 2 where they enter the cell again and reduce hydrogen ions. Electrode 1 is therefore referred to as the minus pole of the cell ("where the electrons come from"). At the anode (electrode 1), the half-reaction is
\(\text{H}_2\text{(g)} \rightarrow 2\text{H}^+\text{(aq)} + 2\text{e}^-\)
In order to calculate the EMF, it is required to calculate the potential difference between both electrodes 1 and 2 compared to the standard hydrogen electrode. The EMF for the cell is then calculated as the sum of reduction potentials
\( \Delta E = E(\text{Kath.}) + E(\text{An.}) \)
and this is where we have to be careful to choose the sign correspondingly to the reaction we are formulating.
For the reaction at the kathode (where reduction of \(\text{H}^+\text{(aq)}\) occurs), the potential (compared to the normal hydrogen electrode) is given by the Nernst equation
\( E\text{(Kath.)} = E^0 - \frac{0.06}{2}\log{Q} \)
The \(2\) in the fraction is the number of transported electrons per reduction equivalent. The \( (-) \)-sign in front of the log-expression is used when the reaction quotient \( Q \) corresponds to a reaction formulated in direction of reduction:
\( 2\text{H}^+ \text{(aq)} + 2\text{e}^- \rightarrow \text{H}_2 \text{(g)} \)
For the general reaction
\( k\cdot K + l\cdot L \rightarrow x\cdot X + y\cdot Y \)
the reaction quotient is always defined as
\( Q = \frac{a^x(X) \cdot a^y(Y)}{a^k(K) \cdot a^l(L)} \),
where the \(a^i\)'s are activities of the species. Here, we understand the hydrogen gas has unit activity, \(a(\text{H}_2\text{(g)}) = 1\), and so
\( E\text{(Kath.)} = E^0 - \frac{0.06}{2}\log \frac{1}{ \left[ \text{H}^+ \right] ^2 } \)
where \( \left[\text{H}^+\right] = 5 \text{M} \) and the exponent \(2\) is the stoichiometric coefficient. We emphasize again the minus sign in front of the \( \log \)-term used when the reaction is formulated in direction of reduction. \(E^0\) is the electrode potential for the normal-hydrogen electrode which is zero by definition.
A similar equation is formulated for the anode half-reaction:
\( E\text{(An.)} = E^0 + \frac{0.06}{2}\log \frac{\left[ \text{H}^+ \right]^2}{1} \)
however here a \((+)\)-sign is used in front of the \(\log\)-term because the reaction is formulated in direction of oxidation (so the oxidized species \(\left[H^+\right]\) appears in the numerator) and \( \left[\text{H}^+\right] = 0.025 \text{M} \).
The numeric values are \(E(\text{Kath.}) = +0.041 V\) and \(E(\text{An.}) = -0.096 V\). \(E(\text{An.})\) however corresponds to an oxidation potential and so in order to calculate the EMF, we need to take its negative (such that we are adding reduction potentials). This then provides us with
\( EMF = \Delta E(\text{Kath.}) + {}^-\Delta E(\text{An.}) = 0.041 + {}^-(-0.096) = 0.137 [V] \)
The question can now be extended by asking what the EMF will be, when the hydrogen gas pressures are different from atmospheric pressure (and so the activity \(a(\text{H}_2)\) is no longer unity):
Assume \(p(\text{H}_2) = 202.6 \text{kPa}\) at the anode and \(p(\text{H}_2) = 10.13\text{kPa}\) at the kathode.
Given these pressures, the activities are \(a(\text{H}_2, \text{Kath.}) = p(\text{H}_2)/101.3 \text{kPa} = 0.1 \) and \(a(\text{H}_2, \text{An.}) = p(\text{H}_2)/101.3 \text{kPa} = 2 \). By substituting these values in the Nernst equations for the respective half-cells (using the same concentrations for \(\left[H^+\right]\)), we arrive at
\( E\text{(Kath.)} = E^0 - \frac{0.06}{2}\log \frac{0.1}{ \left[ \text{H}^+ \right] ^2 } = 0.072 \left[V\right]\)
and
\( E\text{(An.)} = E^0 + \frac{0.06}{2}\log \frac{\left[ \text{H}^+ \right]^2}{2} = -0.105 \left[V\right] .\)
The EMF in this cell is therefore (after again switching the sign of the anode potential to obtain a reduction potential)
\( EMF = \Delta E(\text{Kath.}) + {}^-\Delta E(\text{An.}) = 0.071 + {}^-(-0.105) = 0.177 [V] .\)
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