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%

Governor before election

Charles W. Lippitt
Republican

Elected Governor

Charles W. Lippitt
Republican

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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

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.
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