Heat given off from an electrochemical cell compared to mixing reactants
I think the problem here is related to what each of these terms means, in a fundamental sense.

$\mathrm{d}S$ is the entropy of the system

$\frac{-\mathrm{d}q}{T}$ is the entropy of the surroundings - the heat change of the surroundings divided by the temperature.

$\mathrm{d}H$ is the enthalpy of the system - it is equal to $\mathrm{d}q$ only when the internal energy change occurs at constant pressure, and when there is no form of work being done besides PV work. In an electrochemical cell, moving the electrons is work, and so we can't leave it out of the total energy balance.

In other words, you can't just use $\mathrm{d}q = \Delta H$ for the 1st question because $\mathrm{d}q_p \neq\Delta H$ when you are doing any work besides PV work.

Instead you are using the Clausius inequality, which is just another way of stating the second law: the entropy of the universe must increase for any process. In this case, the universe is the system plus the surroundings, and so:

$$\Delta S - \frac{\mathrm{d}q}{T} \geq 0$$

Solving that for dq gives you the heat absorbed by the surroundings, which is equal to temperature times the entropy change of the system.

For a process that is only (possibly) doing PV work, at constant pressure, $\Delta H = \mathrm{d}q_p$, and so we can just use that. This would be an example of using the 1st law at constant pressure, when there is no other type of work besides PV. 

In your case, you used:
\begin{align}
\Delta G &amp;= \Delta H - T\,\mathrm{d}S\\
\Delta H &amp;= \Delta G + T\,\mathrm{d}S\\
\end{align}

This works because the equation for $\Delta G$ is derived directly from the Clausius inequality at constant pressure. In other words, you are using the 2nd law to derive the same result.

Edit

To answer your last questions, and questions from the comments:


  why is $T\,\mathrm{d}S$ less than $\mathrm{d}q$ in my example? I don't really get this bit


It's not - but you are dropping the signs and that makes it hard to see. If you leave the signs in:

\begin{align}
T\,\mathrm{d}S &amp;\geq \mathrm{d}q\\
(298~\mathrm{K})\left( -61.6~\mathrm{\frac{J}{K\cdot mol}}\right) &amp;\geq -70~\mathrm{\frac{kJ}{mol}}\\
-18~\mathrm{kJ} &amp;\geq -70~\mathrm{kJ}\\
\end{align}

Which is true, although it is easier to see if you divide by -1:

$$18~\mathrm{kJ} \leq 70~\mathrm{kJ}$$


  The entropy divided by the temperature gives $234.9~\mathrm{JK^{−1}mol^{−1}}$ which is not the value for entropy as it should be according to the Clausius inequality given that the change in entropy is equal to the heat given off in part (b). Why is this so?


Here I assume you meant "enthalpy divided by temperature." The second reaction is not carried out in a reversible manner, so we have no guarantee that $\mathrm{d}S = \mathrm{d}q/T$. All we know is that it will be greater than or equal to $\mathrm{d}q/T$, and if you plug in the numbers, it is.

Remember that all of this is built into the $\Delta G$ equation as well, which is why we use it when doing "regular" chemical reactions at constant T &amp; P.