The 1886 Rhode Island gubernatorial election was held on April 7, 1886. Incumbent Republican Governor George P. Wetmore defeated Democratic nominee Amasa Sprague and Prohibition nominee George H. Slade, securing re-election with 53.36% of the popular vote.
Candidates
Three candidates contested the general election:
- George P. Wetmore (Republican) — the incumbent governor, seeking re-election.
- Amasa Sprague (Democratic) — the Democratic nominee.
- George H. Slade (Prohibition) — the Prohibition Party nominee.
Results
| Party | Candidate | Votes | Percentage |
|---|---|---|---|
| Republican | George P. Wetmore (incumbent) | 14,340 | 53.36% |
| Democratic | Amasa Sprague | 9,944 | 37.00% |
| Prohibition | George H. Slade | 2,585 | 9.62% |
| Scattering | — | 6 | 0.02% |
| Total | 26,875 | 100% |
Wetmore's majority over Sprague was 4,396 votes (16.36% of the total). The Republican Party retained control of the governorship. Some sources report Sprague's vote total as 9,994 rather than 9,944.
Context
The election took place during a period when Rhode Island held its gubernatorial elections in the spring, separate from the November federal election cycle. The result was consistent with the Republican Party's dominance in Rhode Island state politics during this era. Contemporary press coverage, such as a report in The New York Times published on April 9, 1886, characterized the outcome as a success for the Republican Party, noting the re-election of Governor Wetmore and his associates on the state ticket, with the exception of the Attorney General.
References
- Dubin, Michael J. (2010). United States Gubernatorial Elections, 1861–1911. Jefferson, NC: McFarland & Co. p. 14.
- Glashan, Roy R. (1979). American Governors and Gubernatorial Elections, 1775–1978. Westport, CT: Meckler Books. pp. 272–273.
- Manual with Rules and Orders for the use of the General Assembly of the State of Rhode Island, 1886–7. Providence, R.I.: E. L. Freeman & Son, State Printers. 1886. pp. 91–92.
