Rhode Island gubernatorial election
1874 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 | Henry Howard | Lyman Pierce | | Party | Republican | Democratic | Popular vote | 12,335 | 1,589 | Percentage | 87.48% | 11.27% | |
Governor before election Henry Howard Republican | Elected Governor Henry Howard Republican | |
The 1874 Rhode Island gubernatorial election was held on 1 April 1874 in order to elect the Governor of Rhode Island. Incumbent Republican Governor Henry Howard won re-election against Democratic nominee Lyman Pierce.[1]
General election
On election day, 1 April 1874, incumbent Republican Governor Henry Howard won re-election by a margin of 10,746 votes against his opponent Democratic nominee Lyman Pierce, thereby retaining Republican control over the office of Governor. Howard was sworn in for his second term on 5 May 1874.[2]
Results
Rhode Island gubernatorial election, 1874 Party | Candidate | Votes | % |
| Republican | Henry Howard (incumbent) | 12,335 | 87.48 |
| Democratic | Lyman Pierce | 1,589 | 11.27 |
| | Scattering | 177 | 1.25 |
Total votes | 14,101 | 100.00 |
| Republican hold |
References