1895 Rhode Island gubernatorial election

The 1895 Rhode Island gubernatorial election was held on April 3, 1895. Republican nominee Charles W. Lippitt defeated Democratic nominee George L. Littlefield with 56.89% of the vote.

1895 Rhode Island gubernatorial election

← 1894April 3, 18951896 →
 
Dem
PRO
NomineeCharles W. LippittGeorge L. LittlefieldSmith Quimby
PartyRepublicanDemocraticProhibition
Popular vote25,09814,2892,624
Percentage56.89%32.39%5.95%

Governor before election

Daniel Russell Brown
Republican

Elected Governor

Charles W. Lippitt
Republican

General election

Candidates

Major party candidates

  • Charles W. Lippitt, Republican
  • George L. Littlefield, Democratic

Other candidates

  • Smith Quimby, Prohibition
  • George Boomer, Socialist Labor
  • William Foster Jr., People's

Results

1895 Rhode Island gubernatorial election[1]
PartyCandidateVotes%±%
RepublicanCharles W. Lippitt 25,098 56.89%
DemocraticGeorge L. Littlefield14,28932.39%
ProhibitionSmith Quimby2,6245.95%
Socialist LaborGeorge Boomer1,7303.92%
PopulistWilliam Foster Jr.3790.86%
Majority10,809
Turnout
Republican holdSwing

References