Reduction potential as a function of pH
Nernst equation is the only way of solving this problem. Let's consider the example of the reduction of permanganate. The half-reaction is:
$$\ce{MnO4-  +  8 H+  +  5 e-  -&gt;  Mn^2+  +  4 H2O}$$

The potential can be written as:

$$E  =  E^\circ  +  \frac{0.059}{5}\log \left(\frac{\ce{[MnO4-][H+]^8}}{\ce{[Mn^2+]}} \right)$$  

If you want to see explicitly the $\mathrm{pH}$, you can rewrite the preceding expression, which gives:

$$E  =  E^\circ  +  0.012 \log \left(\frac{\ce{[MnO4^-]}}{\ce{[Mn^2+]}}\right) - \left(\frac{8\times 0.059}{5} \right) \mathrm{pH}$$

or :
$$E  =  E^\circ  +  0.012 \log \left(\frac{\ce{[MnO4^-]}}{\ce{[Mn^2+]}}\right) - 0.094 \: \mathrm{pH}$$