1891 Iowa gubernatorial election

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.

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

General election

edit

Candidates

edit

Major party candidates

  • Horace Boies, Democratic
  • Hiram 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 (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

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