Rhode Island gubernatorial election
1868 Rhode Island gubernatorial election
|
|
| | | Nominee | Ambrose Burnside | Lyman Pierce | | Party | Republican | Democratic | Popular vote | 10,038 | 3,390 | Percentage | 74.69% | 25.23% | |
Governor before election Ambrose Burnside Republican | Elected Governor Ambrose Burnside Republican | |
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]
General election
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.[2]
Results
Rhode Island gubernatorial election, 1868 Party | Candidate | Votes | % |
| Republican | Ambrose Burnside (incumbent) | 10,038 | 74.69 |
| Democratic | Lyman Pierce | 3,390 | 25.23 |
| | Scattering | 11 | 0.08 |
Total votes | 13,439 | 100.00 |
| Republican hold |
References