# Open circuit voltage

Now that we know how much energy can be released by the reaction at a given temperature, we can find the open circuit voltage of the cell:

For every mole of hydrogen used by the cell, 2 moles of electrons flow from anode to cathode via the external circuit. The charge that flows per a mole of hydrogen consumed / water produced is therefore:

–2N_{A}e = –2F

where N_{A} is the Avogadro’s number, –e is the charge on an electron and F is Faraday’s constant; the charge on a mole of electrons. If E is the voltage of the cell, then the energy required to move this charge around a circuit is:

Electrical work done = charge × voltage = –2FE joules

If the system were totally reversible (i.e. no losses), this work would be equal to \(\Delta {\overline g _f}\) such that:

$$\Delta {\overline g _f} = - 2F \cdot E$$

Which re-arranges to:

$$E = {{ - \Delta {{\overline g }_f}} \over {2F}}$$

And which generalises to:

(9) \(E = \){{ - \Delta {{\overline g }_f}} \over {zF}}\)

where z represents the number of electrons transferred for each mole of fuel. And hence we have an expression that tells us the EMF of the cell (which is the open circuit voltage with no losses). This value assumes no irreversibilities (i.e. 100% efficiency) and that the cell is fuelled by hydrogen and oxygen at standard pressure of 1 atm (0.1 MPa). Using oxygen from the air at the cathode is equivalent to supplying oxygen at a lower pressure. We can take pressure into account using the **Nernst equation**.

We’ve already noted that temperature has an influence on the important \(\Delta {\overline g _f}\) value.

Pressure variation has a more complex, but just as significant effect. Considering a general reaction such that:

$$jJ + kK \to mM$$

Where j moles of J react with k moles of K to give m moles of M. If the reactants are modelled as ideal gases, we describe their activity, a as:

Activity:\(a = \)\({P \over {{P^0}}}\)

where P is the partial pressure of a reactant and P^{0} is standard pressure (0.1 MPa). Activity is linked to concentration in a similar way for dissolved reactants. The activities of reactants and products change the \(\Delta {\overline g _f}\) of the general reaction such that:

$$\Delta {\overline g _f} = \Delta {\overline g _f}^0 - RT\ln \left( {{{{a_J}^j \cdot {a_K}^k} \over {{a_M}^m}}} \right)$$

Which for a hydrogen fuel cell, H_{2} + ½O_{2} → H_{2}O, becomes:

(10) \(\Delta {\overline g _f} = \Delta {\overline g _f}^0 - RT\ln \) \(\left( {{{{a_{{H_2}}} \cdot a_{{O_2}}^{{\textstyle{1 \over 2}}}} \over {{a_{{H_2}O}}}}} \right)\)

Note the trends: as product activity increases, \(\Delta {\overline g _f}\) decreases, i.e. less energy released. As reactant activity increases, \(\Delta {\overline g _f}\) increases hence more energy released. To see how this affects voltage we substitute equation (10) into equation (9) to get:

where E^{0} is the EMF at standard pressure. Equation (11) is an example of a Nernst equation. The EMF calculated using Nernst equations is called a *Nernst Voltage* and is the reversible cell voltage which would exist at a given temperature and pressure.

If, as an example, we took the temperature to be high such as that found in an SOFC, then the H_{2}O could be modelled as an ideal gas meaning that equation (11) becomes:

$$E = {E^0} + {{RT} \over {2F}}\ln \left( {{{{{{P_{{H_2}}}} \over {{P^0}}} \cdot {{\left( {{{{P_{{O_2}}}} \over {{P^0}}}} \right)}^{{\textstyle{1 \over 2}}}}} \over {{{{P_{{H_2}O}}} \over {{P^0}}}}}} \right)$$

And if pressures are given in bar, then standard pressure P^{0} = 1, hence:

(12) $$E = {E^0} + {{RT} \over {2F}}\ln \left( {{{{P_{{H_2}}} \cdot P_{{O_2}}^{{\textstyle{1 \over 2}}}} \over {{P_{{H_2}O}}}}} \right)$$

This is a very useful equation.