1891 Iowa gubernatorial election

1891 Iowa gubernatorial election

← 1889 November 3, 1891 1893 →
 
Nominee Horace Boies Hiram C. Wheeler
Party Democratic Republican
Popular vote 207,594 199,381
Percentage 49.40% 47.45%

County results
Boies:      40-50%      50-60%      60-70%      70-80%
Wheeler:      40-50%      50-60%      60-70%
Westfall:      30-40%

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 Hiram C. Wheeler with 49.40% of the vote.

General election

Candidates

Major party candidates

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

Other candidates

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

Results

1891 Iowa gubernatorial election[1]
Party Candidate Votes % ±%
Democratic Horace Boies (incumbent) 207,594 49.40%
Republican Hiram C. Wheeler 199,381 47.45%
Populist A. J. Westfall 12,303 2.93%
Prohibition Isaac T. Gibson 915 0.22%
Majority 8,213
Turnout
Democratic hold Swing

References

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