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