1864 United States presidential election in Illinois explained

See main article: 1864 United States presidential election.

Election Name:1864 United States presidential election in Illinois
Country:Illinois
Flag Year:1864
Type:presidential
Ongoing:no
Previous Election:1860 United States presidential election in Illinois
Previous Year:1860
Next Election:1868 United States presidential election in Illinois
Next Year:1868
Election Date:November 8, 1864
Turnout:20.34% of the total population 0.50 pp[1]
Image1:Abraham Lincoln November 1863.jpg
Nominee1:Abraham Lincoln
Party1:National Union Party (United States)
Home State1:Illinois
Running Mate1:Andrew Johnson
Electoral Vote1:16
Popular Vote1:189,512
Percentage1:54.42%
Nominee2:George B. McClellan
Party2:Democratic Party (United States)
Home State2:New Jersey
Running Mate2:George H. Pendleton
Electoral Vote2:0
Popular Vote2:158,724
Percentage2:45.58%
Map Size:350px
President
Before Election:Abraham Lincoln
Before Party:Republican Party (United States)
After Election:Abraham Lincoln
After Party:National Union Party (United States)

The 1864 United States presidential election in Illinois took place on November 8, 1864, as part of the 1864 United States presidential election. Illinois voters chose 16 representatives, or electors, to the Electoral College, who voted for president and vice president.[2]

Illinois was won by the National Union candidate incumbent Republican President Abraham Lincoln of Illinois and his running mate former Senator and Military Governor of Tennessee Andrew Johnson. They defeated the Democratic candidate 4th Commanding General of the United States Army George B. McClellan of New Jersey and his running mate Representative George H. Pendleton of Ohio.[2] Lincoln won the state by a margin of 8.84%.

See also

Notes and References

  1. Web site: 1880 Presidential Election Results Illinois Total Population Turnout.
  2. Web site: 1864 Presidential Election Results Illinois.