Eb/No (QAM128) = 1/14σ2 ,,is not correct because in the case of qam you have to consider the constellation of qam before calculating the average power of signal.
It comes out to be correct in case of qpsk because there magnitude is constant, only symbol duration is double of bit duration.
just make the constellation and calculate average power by E(sum(abs(x*x)))..
eg. bpsk Pav= 0.5*1+0.5*abs(-1)=1...........Eb/No (BPSK) = 1/2σ2
qpsk..in your constellation the symbols are at [1,1],[1,-1],[-1,-1],[-1,1]..
Es=2...Eb=1... but here noise have in phase and quadrature components so noice variace = 2σ2....No=4σ2........Eb/No (QPSK) = 1/4σ2