1868 Rhode Island gubernatorial election

The 1868 Rhode Island gubernatorial election was held on 1 April 1868 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 1866 and 1867 election.[1]

1868 Rhode Island gubernatorial election

← 1867 1 April 1868 1869 →
 
Nominee Ambrose Burnside Lyman Pierce
Party Republican Democratic
Popular vote 10,038 5,731
Percentage 63.61% 36.32%

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

Governor before election

Ambrose Burnside
Republican

Elected Governor

Ambrose Burnside
Republican

General election

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On election day, 1 April 1868, incumbent Republican governor Ambrose Burnside won re-election by a margin of 6,648 votes against his opponent Democratic nominee Lyman Pierce, thereby retaining Republican control over the office of Governor. Burnside was sworn in for his third term on 4 May 1868.

Results

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Rhode Island gubernatorial election, 1868[2]
Party Candidate Votes %
Republican Ambrose Burnside (incumbent) 10,038 63.61
Democratic Lyman Pierce 5,731 36.32
Scattering 11 0.07
Total votes 15,780 100.00
Republican hold

References

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  1. ^ "Ambrose Burnside". National Governors Association. Retrieved 7 April 2024.
  2. ^ United States Gubernatorial Elections 1861-1911 (Michael J. Dubin) Page 478