1845 Tennessee gubernatorial election
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| Elections in Tennessee |
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 Government |
The 1845 Tennessee gubernatorial election was held on August 7, 1845, to elect the Governor of Tennessee. Democratic nominee and former member of the U.S. House of Representatives from Tennessee's 6th district Aaron V. Brown defeated Whig nominee and former United States Senator from Tennessee Ephraim H. Foster.[1]
General election[edit]
On election day, 7 August 1845, Democratic nominee Aaron V. Brown won the election by a margin of 1,623 votes against his opponent Whig nominee Ephraim H. Foster, thereby gaining Democratic control over the office of Governor. Brown was sworn in as the 11th Governor of Tennessee on 14 October 1845.[2]
Results[edit]
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Democratic | Aaron V. Brown | 58,269 | 50.71 | |
| Whig | Ephraim H. Foster | 56,646 | 49.29 | |
| Total votes | 114,915 | 100.00 | ||
| Democratic gain from Whig | ||||
References[edit]
- ^ "Gov. Aaron Venable Brown". nga.org. Retrieved 15 February 2024.
- ^ "TN Governor". ourcampaigns.com. 7 June 2005. Retrieved 15 February 2024.
Categories:
- 1845 Tennessee elections
- Tennessee gubernatorial elections
- 1845 United States gubernatorial elections
- August 1845 events
- 1845 in Tennessee
- 1840s in Tennessee
- 1840s Tennessee elections
- 1845 elections
- 1845 elections in North America
- 1845 elections in the United States
- United States gubernatorial elections in the 1840s
- Government of Tennessee