Current and reversible heat in battery reaction

Question

In the book Battery Technology Handbook by Kiehne [1] (Google Books) on page 21, equation (32):

The reversible heat effect per time unit can be related to current flow, because each multiple of the cell reaction requires the current amount $n\cdot F$:"

$$\frac{\mathrm dQ_\mathrm{rev}}{\mathrm dt} = \frac{Q_\mathrm{rev}}{nF}\cdot i \label{eqn:32}\tag{32}$$

with $n$: number of exchanged electrons; $F$: Faraday constant $(\pu{96485 A s/equivalent})$; $i$: current in $\pu{A}$.

Two questions:

1. How is the current related to $nF$? current and $nF$ have different dimensions?

2. How can one derive equation \eqref{eqn:32}?

References

1. Battery Technology Handbook, 2nd ed.; Kiehne, H. A., Ed.; Electrical engineering and electronics; Marcel Dekker: New York, 2003. ISBN 978-0-8247-4249-2.

Update in order to respond to comments

I know that $\Delta G$ is connected with a reversible process and in principle, your derivation makes sense. My problem lies in the formula:

$\Delta S = \frac{Q_{rev}}{T}$ is a basic thermodynamic relation and the Gibbs Energy is defined as $\Delta G = \Delta H - T\Delta S$. So, following your derivation $\Delta H$ would be zero. Since this problem is connected with my other question, which is linked in the comments, I would expect two different heat terms, a reversible one and a irreversible one. And from the formulas there, I expect $Q_{rev}$ to be connected only with entropy. Therefore, $Q_{rev} = -T\frac{\text{d}G}{\text{d}T}$ rather than $Q_{rev} = -\Delta G$ as the combination of your equations (2) and (4) implies.

I is quite hard to explain my problem...

There is another way to get to the equation in question, but this seems to be too simple:

Since, $\frac{\Delta G}{nF} = \text{Voltage}$, $\frac{T\Delta S}{nF} = \text{formal Voltage}$. And $\textit{Voltage}\cdot I = P = \dot{Q} = \frac{\text{d}Q}{\text{d}t}$, equation (32) seems perfectly valid, but this is a kind of a weak derivation.

Answer 1

How is the current related to nF? current and nF have different dimensions?

$nF$ is the charge $C$ transferred during the reaction, while current $i$ is the rate of charge transfer ($dC/dt$).

The following is a derivation, now edited to be more rigorous. The equation you provide is an expression for electrical Joule heating, which follows in the steady-state from the expression for the power (rate of work) generated by the battery. From Ohm's law

$P=\left(\frac{dw_{rev}}{dt}\right)=-i^2R=-i E_0 \tag{1}$

where $E_0$ is the (open-circuit) electric potential, $i$ the current, $w_{rev}$ is the reversible electrical work. Note that $E_0$ is the potential when reversible work is done.

Now assume a steady state, with constant P and T constraints, with work done and heat generated cancelling (so that the internal energy $U = constant$), that is

$Q_{rev} = -w_{rev} \tag{2}$

and therefore also power and rate of heat dissipation equal:

$\left(\frac{dQ_{rev}}{dt}\right)=-\left(\frac{dw_{rev}}{dt}\right)=i E_0 \tag{3}$

Now according to the Nernst equation,

$\Delta G=w_{rev} =-nFE_0 \tag{4}$

In the steady-state the dissipated heat is equal to the electrical work, so that, combining (2) and (4), we have that

$\frac{Q_{rev}}{nF}=-\frac{w_{rev}}{nF}=E_0 \tag{5}$

which leads, combined with (3), to the equation in the book:

$\left(\frac{dQ_{rev}}{dt}\right)=i \frac{Q_{rev}}{nF} \tag{6}$

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Aside

Note I apply the opposite sign convention for work from that in the OP (apologies) - in the convention I use work is positive when performed on the system (charging a battery is positive work, discharging negative work). This does not affect the result of the derivation, since I apply the same sign convention for heat (for an exothermal process heat is negative).

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On $\Delta H$ and an alternative (longer) derivation

For a process at constant p, it is common to encounter the expression

$$\Delta H = Q_p$$

However if there is non-PV work, the more general form of this equation is

$$\Delta H = Q_p + w_{non-pV}$$

This expression is general and applies when the process is carried out either reversibly or irreversibly. In differential form,

$$d\Delta H = dQ_p + dw_{non-pV} \tag{a1}$$

Note also that in the preceding derivation non-pV work is electrical.

When the process is carried out reversibly a maximum amount of work is done and

$$w_{non-pV,rev} =\Delta G = \Delta H - T \Delta S = \Delta H - Q_{rev}$$

This leads to the following expression at constant T, equal to that in the linked problem (but note the different work sign convention here):

$$ d\Delta H = dQ_{rev} + dw_{rev} = d\Delta G + Td\Delta S \tag{a2}$$

where I dropped the "non-pV" subscript since the work is assumed to be electrical.

Since H is a state function it must be equal for reversible and irreversible processes, and we can equate (a1) and (a2), leading to the following general expression:

$$ d\Delta H = dQ_p + dw_{ele}= d\Delta G + Td\Delta S $$

which leads finally to

$$ dQ_p = - dw_{ele} + d\Delta G + Td\Delta S \tag{a3}$$

Taking the time derivative of this equation results in the final expression in the linked problem (making sure to account for differences in the work sign convention):

$$\dot{Q} = \dot{Q}_\text{rev} + \dot{Q}_\text{irrev} = IT\,\frac{\mathrm dE_0}{\mathrm dT} + I(E-E_0) \tag{a4}$$

To obtain the equation in this problem it is only necessary to apply the reversibility condition starting from either (a3) or (a4). Starting from (a3),

$$ dQ_{rev} = - dw_{ele,rev} + d\Delta G + Td\Delta S $$

But $ dw_{ele,rev} = d\Delta G $ which leads to (somewhat trivially)

$$ dQ_{rev} = Td\Delta S $$

Taking the time derivative and inserting the Nernst expression for $\Delta S$ (twice!) gives

$$\dot{Q}_\text{rev} = \left(\frac{dQ_{rev}}{dt}\right) = IT\,\frac{\mathrm dE_0}{\mathrm dT} = I\frac{Q_{rev}}{nF}$$

which is the desired expression. The same result can be arrived at by applying the reversibility condition to equation (a4).