1868 Maine gubernatorial election explained

Election Name:1868 Maine gubernatorial election
Image1:File:Chamberlain_as_Governor_of_Maine.jpg
Before Election:Joshua Chamberlain
Governor
Percentage2:42.67%
Popular Vote2:56,207
Party2:Democratic Party (United States)
Nominee2:Eben F. Pillsbury
Popular Vote1:75,523
Country:Maine
Percentage1:57.33%
Party1:Republican Party (United States)
Nominee1:Joshua Chamberlain
Map Size:300px
Election Date:September 14, 1868
Next Year:1869
Next Election:1869 Maine gubernatorial election
Previous Year:1867
Previous Election:1867 Maine gubernatorial election
Ongoing:No
Type:presidential
After Party:Republican Party (United States)
After Election:Joshua Chamberlain
Before Party:Republican Party (United States)

The 1868 Maine gubernatorial election was held on September 14, 1868. Incumbent Republican Governor and war hero Joshua Chamberlain defeated the Democratic candidate Eben F. Pillsbury.[1] [2]

General election

Candidates

Republican

Democratic

Results

Chamberlain won reelection to a third term, and won a majority of 19,316 votes.

References

  1. Book: Dubin, Michael J.. United States Gubernatorial Elections, 1861_ÑÐ1911: The Official Results by State and County. 2010-06-11. McFarland. 978-0-7864-5646-8. en.
  2. Web site: Our Campaigns - ME Governor Race - Sep 14, 1868. 2021-09-01. www.ourcampaigns.com.