1852 United States presidential election in Georgia explained

See main article: 1852 United States presidential election.

Election Name:1852 United States presidential election in Georgia
Country:Georgia (U.S. state)
Flag Year:1852
Type:presidential
Ongoing:no
Previous Election:1848 United States presidential election in Georgia
Previous Year:1848
Next Election:1856 United States presidential election in Georgia
Next Year:1856
Election Date:November 2, 1852
Image1:Mathew Brady - Franklin Pierce (cropped).jpg
Nominee1:Franklin Pierce
Party1:Democratic Party (United States)
Home State1:New Hampshire
Running Mate1:William R. King
Electoral Vote1:10
Popular Vote1:40,516
Percentage1:64.70%
Nominee2:Winfield Scott
Party2:Whig Party (United States)
Home State2:New Jersey
Running Mate2:William A. Graham
Electoral Vote2:0
Popular Vote2:16,660
Percentage2:26.60%
Image3:Daniel Webster crop.jpg
Nominee3:Daniel Webster
Party3:Union
Color3:FFAADD
Home State3:Massachusetts
Running Mate3:Charles J. Jenkins
Electoral Vote3:0
Popular Vote3:5,324
Percentage3:8.50%
Map Size:290px
President
Before Election:Millard Fillmore
Before Party:Whig Party (United States)
After Election:Franklin Pierce
After Party:Democratic Party (United States)

The 1852 United States presidential election in Georgia took place on November 2, 1852, as part of the 1852 United States presidential election. Voters chose 10 representatives, or electors to the Electoral College, who voted for President and Vice President.

Georgia voted for the Democratic candidate, Franklin Pierce, over Commanding General Winfield Scott, the nominee of the Whig Party, and Senator Daniel Webster. Having been denied the Whig nomination at the party's 1852 National Convention, Webster was placed on the ballot without permission by one faction of the state's Constitutional Union Party, but died of natural causes shortly before the election.[1] Pierce won Georgia by a margin of 38.10%.

Notes and References

  1. Ogg, Frederic Austin (1914). Daniel Webster. p. 407.