1867 Rhode Island gubernatorial election

The 1867 Rhode Island gubernatorial election was held on 3 April 1867 in order to elect the governor of Rhode Island. Incumbent Republican governor Ambrose Burnside won re-election against Democratic nominee Lyman Pierce in a rematch of the previous election.[1]

1867 Rhode Island gubernatorial election

← 1866 3 April 1867 1868 →
 
Nominee Ambrose Burnside Lyman Pierce
Party Republican Democratic
Popular vote 7,372 3,178
Percentage 69.84% 30.11%

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

Governor before election

Ambrose Burnside
Republican

Elected Governor

Ambrose Burnside
Republican

General election

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On election day, 3 April 1867, incumbent Republican governor Ambrose Burnside won re-election by a margin of 4,194 votes against his opponent Democratic nominee Lyman Pierce, thereby retaining Republican control over the office of governor. Burnside was sworn in for his second term on 4 May 1867.[2]

Results

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Rhode Island gubernatorial election, 1867
Party Candidate Votes %
Republican Ambrose Burnside (incumbent) 7,372 69.84
Democratic Lyman Pierce 3,178 30.11
Scattering 6 0.05
Total votes 10,556 100.00
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.