Rhode Island gubernatorial election
1867 Rhode Island gubernatorial election![](//upload.wikimedia.org/wikipedia/commons/thumb/f/f3/Flag_of_Rhode_Island.svg/50px-Flag_of_Rhode_Island.svg.png)
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| | | Nominee | Ambrose Burnside | Lyman Pierce | | Party | Republican | Democratic | Popular vote | 7,372 | 3,178 | Percentage | 69.84% | 30.11% | |
Governor before election Ambrose Burnside Republican | Elected Governor Ambrose Burnside Republican | |
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]
General election
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
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