1866 Rhode Island gubernatorial election

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1866 Rhode Island gubernatorial election
Flag of the United States (1865-1867).svg
  1865 4 April 1866 1867  
  Ambrose Burnside - retouched.jpg No image.svg
Nominee Ambrose Burnside Lyman Pierce
Party Republican Democratic
Popular vote8,1972,816
Percentage73.36%25.20%

1866 Rhode Island gubernatorial election results map by county.svg
County results
Burnside:      60–70%     70–80%     80–90%

Governor before election

James Y. Smith
Republican

Elected Governor

Ambrose Burnside
Republican

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]

Contents

General election

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

Rhode Island gubernatorial election, 1866
PartyCandidateVotes%
Republican Ambrose Burnside 8,197 73.36
Democratic Lyman Pierce2,81625.20
Scattering1601.44
Total votes11,221 100.00
Republican hold

References

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