1896 Rhode Island gubernatorial election

1896 Rhode Island gubernatorial election

←  1895 April 1, 1896 1897  →

Dem PRO Nominee Charles W. Lippitt George L. Littlefield Thomas H. Peabody Party Republican Democratic Prohibition Popular vote 28,472 17,061 2,950 Percentage 56.40% 33.79% 5.84%

County results Lippitt:      50–60%     60–70%

Governor before election Charles W. Lippitt Republican Elected Governor Charles W. Lippitt Republican

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

Candidates

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

1896 Rhode Island gubernatorial election ^[1]

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

1. Moore, John Leo, ed. (1994). Congressional Quarterly's Guide to U.S. elections. CQ Press. ISBN   9780871879967 . Retrieved July 18, 2020.