1866 Rhode Island gubernatorial election

1866 Rhode Island gubernatorial election

← 1865 April 4, 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

The 1866 Rhode Island gubernatorial election was held on April 4, 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]

General election

On election day, April 4, 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 May 1, 1866.^[2]

Results

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

1. "Ambrose Burnside". National Governors Association. Retrieved 2024-04-07.
2. "RI Governor". ourcampaigns.com. July 26, 2005. Retrieved 2024-04-07.