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:

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)
\( EMF = \Delta E(\text{Kath.}) + {}^-\Delta E(\text{An.}) = 0.071 + {}^-(-0.105) = 0.177 [V] .\)