The effect on the reduction potential of the half-cell ${\mathrm{MnO}}_4^{-},{\mathrm{Mn}}^{2+},{\mathrm{H}}^{+}\mid \mathrm{Pt}$ on increasing pH of the solution is: 

The reduction reaction is ${\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$ The half-cell potential is

$E=E^{\circ }-\frac{RT}{5F}\ln \left(\frac{\left[{\mathrm{Mn}}^{2+}\right]}{\left[{\mathrm{MnO}}_4^{-}\right]{\left[{\mathrm{H}}^{+}\right]}^5}\right)$

Only numerical values of terms within the brackets are to be used. It pH of the solution is increased, [H+] will decrease and the second term on the right-hand side of the above reaction will become more negative. Hence, the value of E will decrease.