1896 Rhode Island gubernatorial election

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.

1896 Rhode Island gubernatorial election
April 1, 1896
 
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%
Governor before election

Charles W. Lippitt
Republican

 Elected Governor

Charles W. Lippitt
Republican

Elections in Rhode Island
Federal government
U.S. President
U.S. Senate
U.S. House
State government
General elections
Gubernatorial elections
State Senate elections
State House of Representatives elections
Providence
Mayoral elections

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 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%
People's 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.
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