1865 Iowa gubernatorial election explained

Election Name:1865 Iowa gubernatorial election
Country:Iowa
Type:Presidential
Ongoing:no
Previous Election:1863 Iowa gubernatorial election
Previous Year:1863
Next Election:1867 Iowa gubernatorial election
Next Year:1867
Election Date:10 October 1865
Nominee1:William M. Stone
Party1:Republican Party (United States)
Popular Vote1:70,461
Percentage1:56.41%
Nominee2:Thomas H. Benton
Party2:Democratic Party (United States)
Popular Vote2:54,090
Percentage2:43.31%
Map Size:240px
Governor
Before Election:William M. Stone
Before Party:Republican Party (United States)
After Election:William M. Stone
After Party:Republican Party (United States)

The 1865 Iowa gubernatorial election was held on 10 October 1865 in order to elect the Governor of Iowa. Incumbent Republican Governor William M. Stone was re-elected against Democratic nominee Thomas H. Benton.[1]

General election

On election day, 10 October 1865, incumbent Republican Governor William M. Stone won re-election by a margin of 16,371 votes against his opponent Democratic nominee Thomas H. Benton, thereby holding Republican control over the office of Governor. Stone was sworn in for his second term on 16 January 1866.[2]

Results

Notes and References

  1. Web site: William Milo Stone . 28 November 2023 . National Governors Association.
  2. Web site: IA Governor . ourcampaigns.com . 25 July 2005 . 28 November 2023.