Question

17-16. The sensitivity of a coulometer is governed by the delivery of its minimum current for its minimum time. Suppose thatcan be delivered for 0. 1s.

(a) How many moles of electrons are delivered by 5mAfor 0. 1s ?

(b) How many millilitres of a 0.01 Msolution of a two-electron reducing agent is required to deliver the same number of electrons?

Short answer

$a)n(e^{-})=5.18.{10}^{-9}molb)V=2.59.{10}^{-4}mL$

Step-by-step solution

Mole of electrons:

1 mole of electrons holds the Avogadro constant, L, electrons - that is, electrons. You would also be provided that in an exam if you needed to use it. This is known as the Faraday constant.

The sensitivity of a coulometer:

The delivery of a coulometer's minimal current during its minimum time determines its sensitivity. Assume that 5 mA can be given for 0.1 second

Find the mole of electrons:

a) First we need to calculate the charge :

$$\begin{gathered} \mathrm{q}=\mathrm{l}.\mathrm{t}=5.{10}^{-3}\mathrm{A}.0.1\mathrm{s} \\ \mathrm{q}=5.{10}^{-4}\mathrm{C} \end{gathered}$$

So, the moles of electrons that are delivered are:

$$\begin{gathered} \mathrm{n}\left({\mathrm{e}}^{-}\right)=\frac{\mathrm{q}}{\mathrm{F}}=\frac{5.{10}^{-4}\mathrm{C}}{96485\mathrm{C}/\mathrm{mol}} \\ \mathrm{n}\left({\mathrm{e}}^{-}\right)=5.18.{10}^{-9}\mathrm{mol} \end{gathered}$$

Find the sensitivity of a coulometer:

b) A 0.01M solution of a two-electron reducing agent deliver 0.02M electrons. So, the volume is:

$$\begin{gathered} \mathrm{V}=\frac{\mathrm{n}\left({\mathrm{e}}^{-}\right)}{\mathrm{c}}=\frac{5.18.{10}^{-9}\mathrm{mol}}{0.02\mathrm{M}} \\ \mathrm{V}=2.59.{10}^{-7}\mathrm{L} \\ \mathrm{V}=2.59.{10}^{-4}\mathrm{mL} \end{gathered}$$