1866 Rhode Island gubernatorial election

The 1866 Rhode Island gubernatorial election was held on 4 April 1866 in order to elect the governor of Rhode Island. Republican nominee and former Union Army Major General Ambrose Burnside defeated Democratic nominee Lyman Pierce.[1]

1866 Rhode Island gubernatorial election

← 1865 4 April 1866 1867 →
 
Nominee Ambrose Burnside Lyman Pierce
Party Republican Democratic
Popular vote 8,197 2,816
Percentage 73.36% 25.20%

County results
Burnside:      60–70%      70–80%      80–90%

Governor before election

James Y. Smith
Republican

Elected Governor

Ambrose Burnside
Republican

General election

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On election day, 4 April 1866, Republican nominee Ambrose Burnside won the election by a margin of 5,381 votes against his opponent Democratic nominee Lyman Pierce, thereby retaining Republican control over the office of governor. Burnside was sworn in as the 30th governor of Rhode Island on 1 May 1866.[2]

Results

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Rhode Island gubernatorial election, 1866
Party Candidate Votes %
Republican Ambrose Burnside 8,197 73.36
Democratic Lyman Pierce 2,816 25.20
Scattering 160 1.44
Total votes 11,221 100.00
Republican hold

References

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  1. ^ "Ambrose Burnside". National Governors Association. Retrieved 7 April 2024.
  2. ^ "RI Governor". ourcampaigns.com. 26 July 2005. Retrieved 7 April 2024.