1886 Rhode Island gubernatorial election

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1886 Rhode Island gubernatorial election

← 1885 April 7, 1886 1887 →
 
Nominee George P. Wetmore Amasa Sprague George H. Slade
Party Republican Democratic Prohibition
Popular vote 14,340 9,944 2,585
Percentage 53.36% 37.00% 9.62%

Governor before election

George P. Wetmore
Republican

Elected Governor

George P. Wetmore
Republican

The 1886 Rhode Island gubernatorial election was held on April 7, 1886. Incumbent Republican George P. Wetmore defeated Democratic nominee Amasa Sprague with 53.36% of the vote.

General election

Candidates

Major party candidates

Other candidates

  • George H. Slade, Prohibition

Results

1886 Rhode Island gubernatorial election[1][2][3][4][5][a]
Party Candidate Votes % ±%
Republican George P. Wetmore (incumbent) 14,340 53.36%
Democratic Amasa Sprague 9,944 37.00%
Prohibition George H. Slade 2,585 9.62%
Scattering 6 0.02%
Majority 4,396 16.36%
Turnout 26,875
Republican hold Swing

Notes

  1. ^ Some sources gives Sprague's vote as 9,994.[6][7]

References

  1. .
  2. .
  3. .
  4. ^ McPherson, Edward, ed. (1887). The Tribune Almanac and Political Register for 1887. New York, NY: The Tribune Association. p. 79.
  5. ^ Manual with Rules and Orders for the use of the General Assembly of the State of Rhode Island. 1886-7. State of Rhode Island manual. Providence, R. I.: E. L. Freeman & Son, State Printers. 1886. pp. 91–92.
  6. ^ "RI Governor, 1886". Our Campaigns. Retrieved September 14, 2021.
  7. . Retrieved October 14, 2020.