1891 Iowa gubernatorial election

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1891 Iowa gubernatorial election

← 1889 November 3, 1891 1893 →
  Horace Boies signed.jpg No image.svg
Nominee Horace Boies Herman C. Wheeler
Party Democratic Republican
Popular vote 207,594 199,381
Percentage 49.40% 47.45%

Governor before election

Horace Boies
Democratic

Elected Governor

Horace Boies
Democratic

The 1891 Iowa gubernatorial election was held on November 3, 1891. Incumbent Democrat Horace Boies defeated Republican nominee Herman C. Wheeler with 49.40% of the vote.

General election[edit]

Candidates[edit]

Major party candidates

  • Horace Boies, Democratic
  • Herman C. Wheeler, Republican

Other candidates

  • A. J. Westfall, People's
  • Isaac T. Gibson, Prohibition

Results[edit]

1891 Iowa gubernatorial election[1]
Party Candidate Votes % ±%
Democratic Horace Boies 207,594 49.40%
Republican Herman C. Wheeler 199,381 47.45%
People's A. J. Westfall 12,303 2.93%
Prohibition Isaac T. Gibson 915 0.22%
Majority 8,213
Turnout
Democratic hold Swing

References[edit]

  1. ^ Kalb, Deborah (24 December 2015). Guide to U.S. Elections. ISBN 9781483380353. Retrieved August 9, 2020.