1896 Rhode Island gubernatorial election
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The 1896 Rhode Island gubernatorial election was held on April 1, 1896. Incumbent Republican Charles W. Lippitt defeated Democratic nominee George L. Littlefield with 56.40% of the vote.
General election
[edit]Candidates
[edit]Major party candidates
- Charles W. Lippitt, Republican
- George L. Littlefield, Democratic
Other candidates
- Thomas H. Peabody, Prohibition
- Edward W. Theinert, Socialist Labor
- Henry A. Burlingame, People's
Results
[edit]| Party | Candidate | Votes | % | ±% | |
|---|---|---|---|---|---|
| Republican | Charles W. Lippitt (incumbent) | 28,472 | 56.40% | ||
| Democratic | George L. Littlefield | 17,061 | 33.79% | ||
| Prohibition | Thomas H. Peabody | 2,950 | 5.84% | ||
| Socialist Labor | Edward W. Theinert | 1,272 | 2.52% | ||
| Populist | Henry A. Burlingame | 730 | 1.45% | ||
| Majority | 11,411 | ||||
| Turnout | |||||
| Republican hold | Swing | ||||
References
[edit]- ^ Moore, John Leo, ed. (1994). Congressional Quarterly's Guide to U.S. elections. CQ Press. ISBN 9780871879967. Retrieved July 18, 2020.