An oxidation-reduction reaction in which 3 electrons are transferred has a $$\Delta G^0$$ of $$17.37\,\text{kJ mol}^{-1}$$ at $$25^\circ\text{C}$$. The value of $$E^0_{\text{cell}}$$ (in V) is __________ $$\times 10^{-2}$$. ($$1\text{F} = 96500\,\text{C mol}^{-1}$$)

For an electrochemical cell under standard conditions, the fundamental relation between the standard Gibbs free-energy change and the standard cell potential is given first:

$$\Delta G^{0} = -\,n F E^{0}_{\text{cell}}$$

Here, $$\Delta G^{0}$$ is the standard Gibbs free-energy change, $$n$$ is the number of electrons transferred in the balanced redox reaction, $$F$$ is the Faraday constant, and $$E^{0}_{\text{cell}}$$ is the standard cell potential that we wish to determine.

We have $$\Delta G^{0} = 17.37\,\text{kJ mol}^{-1}$$. Converting this to joules because the Faraday constant is in coulombs per mole, we write

$$\Delta G^{0} = 17.37\,\text{kJ mol}^{-1} \times 1000\,\dfrac{\text{J}}{\text{kJ}} = 17370\,\text{J mol}^{-1}.$$

The number of electrons transferred is given as $$n = 3$$, and the Faraday constant is $$F = 96500\,\text{C mol}^{-1}.$$

Now we substitute every known value into the relation $$\Delta G^{0} = -\,n F E^{0}_{\text{cell}}$$ and solve for $$E^{0}_{\text{cell}}$$:

$$17370 = -\,\bigl(3\bigr)\bigl(96500\bigr) \, E^{0}_{\text{cell}}.$$

First multiply $$n$$ and $$F$$:

$$3 \times 96500 = 289500.$$

So the equation becomes

$$17370 = -\,289500 \, E^{0}_{\text{cell}}.$$

Now isolate $$E^{0}_{\text{cell}}$$ by dividing both sides by $$-289500$$:

$$E^{0}_{\text{cell}} = -\,\dfrac{17370}{289500}.$$

Carrying out the division, we note that $$289500 \times 0.06 = 17370$$ exactly, so

$$E^{0}_{\text{cell}} = -\,0.06\,\text{V}.$$

The question requests the answer in the form $$\text{(number)} \times 10^{-2}$$ volts. Writing $$-0.06\,\text{V}$$ as $$-6 \times 10^{-2}\,\text{V},$$ we identify the numerical coefficient as $$-6$$.

So, the answer is $$-6$$.