Am I using the Nernst equation correctly?
I will show you why $\ce{0.295}$ is correct. :-)



Nernst equation can be written in different ways, I let you choose which you prefer. So for any reaction such as $$\ce{\nu_{Ox}Ox + n e^- \rightleftharpoons  \nu_{Red}Red}$$

One can show that at $\ce{25°C}$,

$$E_{Ox/Red} = E_{Ox/Red}^{°} + \frac{RT}{nF}\ln\left(\frac{a_{Ox}^{\nu_{Ox}}}{a_{Red}^{\nu_{Red}}}\right) \tag 1$$

Which can also be written as, see after how, 

$$E_{Ox/Red} = E_{Ox/Red}^{°} + \frac{0.06}{n}\log\left(\frac{a_{Ox}^{\nu_{Ox}}}{a_{Red}^{\nu_{Red}}}\right) \tag 2$$

Where $\log$ is the logarithm in base $\ce{10}$.



So we have $$\frac{RT}{F}\ce{=\frac{8.314 \times 298}{96500}}$$
And we know that $\log(x)=\ln(x)/\ln(10)$ then the coefficient next to $\log$ is $$\frac{RT}{F}\times \ln(10)=\frac{8.314 \times 298}{96500} \times \ln(10) = 0.0591 \approx 0.06$$

To be a little more perfect I sometimes use $\ce{0.059}$ instead of $\ce{0.06}$



So for two electrons exchange the factor coefficient is $\ce{0.059/2}=0.295$