Pearson's $\chi^{2}$ value? Pearson's $\chi^{2}$ value is 115.778 and 4 cells (33.33%) have expected count has less than 5. I am confused how to interpret this output. I have a 2 by 6 contingency table. Can someone tell me what has gone wrong?
 A: Why do you say something has 'gone wrong'? Everything looks fine there to me.
The usual "expected greater than 5" rule of thumb (well over 60 years old now, I believe) is too strict. But if you have a number of values with expected around 1 or less, especially if all the expected values aren't close to equal, then you may have some cause to worry about using the chi-square approximation.
Even so, while the chi-square approximation to the distribution of the test statistic under the null may not be terribly close to the actual (discrete) distribution, it simply means that your p-value won't be as accurate as we might like. 
It would be possible to get an accurate p-value via simulation (this is simple in R for example, where it's built into the chisq.test function).
[There's an example of me using this facility here - done first using the chi-square approximation, which gave a p-value of 0.00096 and then using simulation, which gave a p-value of 0.00032. In your case, I think your p-values will be much lower still.]
Nevertheless, it's not particularly necessary because your chi-square value is so large; if your significance level is any of the common ones (like 5% or 1% say, or even a good deal smaller), you'd still reject the null very comfortably.
I think the only reason to bother computing an accurate p-value would be to appease some journal editor or something like that. Unless it's reasonably easy for you to obtain the simulated distribution of the test statistic (or a simulated p-value) I'd be saying "the null is obviously rejected, we're done".
